UNIVERSIDAD COMPLUTENSE DE MADRID FACULTAD DE CIENCIAS MATEMÁTICAS DEPARTAMENTO DE ANÁLISIS MATEMÁTICO TESIS DOCTORAL Analytical techniques on multilinear problems Técnicas analíticas en problemas multilineales MEMORIA PARA OPTAR AL GRADO DE DOCTOR PRESENTADA POR Daniel Cariello DIRECTOR Juan Benigno Seoane Sepúlveda Madrid, 2017 © Daniel Cariello, 2016 Universidad Complutense de Madrid Facultad de Ciencias Matemáticas Departamento de Análisis Matemático Analytical Techniques on Multilinear Problems (Técnicas Analíticas en Problemas Multilineales) Memoria para optar al grado de doctor presentada por Daniel Cariello Bajo la dirección del doctor Juan Benigno Seoane Sepúlveda Madrid, 2016 ISBN:XXXXXX O esforço é grande e o homem é pequeno. Eu, Diogo Cão, navegador, deixei Este padrão ao pé do areal moreno E para adiante naveguei. A alma é divina e a obra é imperfeita. Este padrão assinala ao vento e aos céus Que, da obra ousada, é minha a parte feita: O por-fazer é só com Deus. E ao imenso e possível oceano Ensinam estas Quinas, que aqui vês, Que o mar com fim será grego ou romano: O mar sem fim é português. E a Cruz ao alto diz que o que me há na alma E faz a febre em mim de navegar Só encontrará de Deus na eterna calma O porto sempre por achar. Fernando Pessoa – Padrão Agradecimientos Existem muitas pessoas que tornaram esse trabalho possível, mas nenhuma tão importante quanto a minha esposa Fernanda. Eu te agradeço por todo carinho, ajuda e paciência. Desejo que você seja feliz e que me veja sempre simples e sorridente ao seu lado. Esse trabalho também não existiria sem o apoio dos meus pais e da minha irmã. Muito obrigado Angela, Sergio e Raquel, que a nossa família cresça e que sejamos felizes juntos. Agradeço também a família da Fernanda que sempre nos ajudou muito. Muito obrigado Francisca, Tsuneo, Fábio e Flávia. Com respeito a minha pesquisa, eu só gostaria que ela estivesse a altura do apoio que recebi da faculdade de matemática da Universidade Federal de Uberlândia, principalmente dos amigos Germano, Vinícius e Geraldo. A oportunidade de expandir meu trabalho só foi possível graças a nossa colaboração e amizade. Me gustaría agradecer a la Universidad Complutense de Madrid por todo el apoio y ayuda. En especial, agradezco a mi director de tesis profesor Dr. Juan Benigno Seoane Sepúlveda por toda la ayuda y principalmente por darme problemas muy interesantes. Deseo que siga investigando con mucho gusto y que siga teniendo mucho éxito. Por fim meus agradecimentos ao governo brasileiro. A colaboração com pesquisadores es­ trangeiros é importantíssima. Algumas ideias que tenho hoje são fruto de conversas que tive com pesquisadores extremamente talentosos. Desejo compartilhá-las com qualquer um que tenha interesse nos mesmos problemas. Agradeço ao apoio financeiro do CNPq, Conselho Na­ cional de Desenvolvimento Cientiífico e Tecnológico - Brasil. Processo número 245277/2012-9. Sobre esta Tesis El desarrollo de esta tesis ha sido posible gracias a la Beca 245277/2012-9 concedida por el Conselho Nacional de Desenvolvimento Científico e Tecnológico - Brasil. Fruto del trabajo de estos cuatro años han sido los artículos: 1. G. Botelho, D. Cariello, V.V. Fávaro, D. Pellegrino, J.B. Seoane-Sepúlveda. On very non-linear subsets of continuous functions. The Quarterly Journal of Mathematics. 65 (2014), 841–850. 2. D. Cariello, J.B. Seoane-Sepúlveda. Basic sequences and spaceability in �p spaces. Journal of Functional Analysis. 266 (2014), 3797–3814. 3. D. Cariello. Separability for weakly irreducible matrices. Quantum Inf. Comp. 14 (2014), 1308–1337. 4. D. Cariello. Does symmetry imply PPT property? Quantum Inf. Comp. 15 (2015), 812–824. 5. D. Cariello. Completely Reducible Maps in Quantum Information Theory. IEEE Transactions on Information Theory 62 (2016) 1721–1732. Contents Resumen iii Abstract v Introducción 1 1 Introduction 9 2 Some Results from the Perron-Frobenius Theory 17 2.1 Two theorems from the Perron-Frobenius Theory . . . . . . . . . . . . . . . . . 18 2.2 Two related properties: completely reducibility and the decomposition property 21 3 Completely Reducible Maps in Quantum Information Theory 25 3.1 Main Theorems: The Complete Reducibility of FA ○GA :Mk →Mk . . . . . . . 26 3.1.1 Main Theorem for PPT matrices . . . . . . . . . . . . . . . . . . . . . . . 26 3.1.2 Main Theorems for SPC matrices and Matrices Invariant under Re­ alignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.2 Applications to Quantum Information Theory . . . . . . . . . . . . . . . . . . . 31 3.2.1 The Separability Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.2.2 Extension of Mutually Unbiased Bases . . . . . . . . . . . . . . . . . . . 35 3.3 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.3.1 Some Remarks on our Main Theorems . . . . . . . . . . . . . . . . . . . 38 3.3.2 A Remark on the Application to the Separability Problem . . . . . . . 41 i ii Contents 4 An application of Borsuk-Ulam Theorem 47 4.1 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.2 An Infinite Dimensional Example . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.3 Open Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 5 Basic Sequences in �p spaces 53 5.1 The case: X = �p, p ∈ [1, ∞[ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 5.2 The case: X = c0 or �∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Resumen Esta tesis doctoral se centra principalmente en tres problemas multilineales y su objetivo es describir las técnicas analíticas y topológicas útiles para atacar estos problemas. El primer problema tiene su origen en la Teoría de Información Cuántica, es el llamado problema de la separabilidad de los estados cuánticos, y los otros dos fueron propuestos por Vladimir I. Gurariy. Denotemos por Mk al conjunto de las matrices complejas de orden k y Pk será el conjunto de matrices Hermíticas semidefinidas positivas de Mk. El objetivo de nuestro primer problema es encontrar un criterio determinístico para distinguir los estados separables de los estados entrelazados. Aqui sólo trabajamos con el caso bipartito de dimensión finita, luego los estados son los elementos del producto tensorial Mk ⊗Mm. Decimos que B ∈ Mk ⊗Mm es separable si B = ∑n i=1 Ci ⊗Di, donde Ci ∈ Pk y Di ∈ Pm, para cada i. Si B no es separable entonces B está entrelazada. Sea VMkV el conjunto {V XV,X ∈ Mk}, donde V ∈ Mk es una proyección ortogonal. Se dice que una transformación lineal T : VMkV → WMmW es una aplicación positiva, si T (Pk ∩ VMkV ) ⊂ Pm ∩WMmW . Se dice que una aplicación no nula positiva T : VMkV → VMkV es irreducible si V ′ MkV ′ ⊂ VMkV es tal que T (V ′ MkV ′) ⊂ V ′ MkV ′ entonces V ′ = V o V ′ = 0. Digamos que T : VMkV → VMkV es una aplicación completamente reducible, si es positiva y si hay proyecciones ortogonales V1, . . . , Vs ∈ Mk tales que ViVj = 0 (i ≠ j), ViV = Vi (1 ≤ i ≤ s), VMkV = V1MkV1 ⊕. . .⊕VsMkVs ⊕R, R ⊥ V1MkV1 ⊕. . .⊕VsMkVs y que satisfacen: T (ViMkVi) ⊂ ViMkVi (1 ≤ i ≤ s), T �ViMkVi es irreducible (1 ≤ i ≤ s), T �R ≡ 0. Sea A = ∑n i=1 Ai ⊗ Bi ∈ Mk ⊗Mm. Defina GA : Mk → Mm, GA(X) = ∑n i=1 tr(AiX)Bi y FA : Mm →Mk, FA(X) = ∑i n =1 tr(BiX)Ai. Nuestros resultados principales son los seguintes: Si A es positiva bajo transposición parcial (PPT) o simétrica con coeficientes positivos (SPC) o invariante bajo realineamiento luego FA ○GA :Mk →Mk es completamente reducible. iii Podemos reducir el problema de la separabilidad de los estados cuánticos para el caso PPT débilmente irreducible, utilizando ese teorema. Otra de las aplicaciones de nuestros resultados principales es la siguiente: Si FA ○ GA : Mk → Mk es completamente reducible con los únicos valores propios 1 o 0 entonces A es separable. El uso de este teorema para una matriz invariante bajo realineamiento, proporciona una prueba diferente del siguiente resultado publicado recientemente: Si hay k bases mutuamente imparciales en Ck entonces existe otra base ortonormal que es mutuamente imparcial con estas k bases. Por lo tanto, si Ck contiene k bases mutuamente imparciales entonces Ck contiene k + 1. El caso real sigue de manera análoga: Si R2k contiene k bases mutuamente imparciales entonces R2k contiene k + 1. Nuestro segundo problema fue propuesto originalmente por Vladimir I. Gurariy y, más tarde, estudiado por Gurariy y Quarta. Sea K un espacio topológico. Consideremos C(K) el espacio vectorial de las funciones reales continuas con dominio K. Denotemos por C�(K) el subconjunto de C(K) formado por aquellas funciones que alcanzan su máximo en un solo punto de K. El conjunto C�(K) no es un espacio vectorial por muchas razones, por ejemplo, la función cero no es un elemento de este conjunto. Gurariy y Quarta se hicieron la siguiente pregunta: ¿Podemos encontrar un subespacio V de C(K) dentro de C�(K) ∪ {0}? En caso positivo, ¿cómo de grande puede llegar a ser la dimensión de V ? Hemos demostrado que si K es un subconjunto compacto de Rn y si V es un subespacio de C(K) dentro C�(K)∪{0} entonces dim(V ) ≤ n. Ese teorema sigue del teorema de Borsuk- Ulam. Nuestro tercer y último problema fue propuesto por Richard M. Aron Y Vladimir I. Gurariy. ¿Es posible obtener un subespacio cerrado de dimensión infinita de �∞ tal que cada sucesión de este espacio tiene solamente un número finito de coordenadas nulas? Si X representa c0 o �p, p ∈ [1, +∞], designaremos por Z(X) el subconjunto de X formado por sucesiones que tienen sólo un número finito de coordenadas cero. Hemos obtenido la respuesta definitiva a esta pregunta. Es decir, (i) No hay subespacio cerrado de dimensión infinita de X dentro Z(X) ∪ {0}. (ii) Hay un subespacio cerrado de dimensión infinita de X dentro V ∖Z(V ), para cualquier subespacio cerrado V (con dimensión infinita) de X. Para obtener estos resultados, se construyen sucesiones básicas dentro de cualquier sube­ spacio cerrado de dimensión infinita de X, y verificando propiedades “especiales”. Una de estas propiedades es que cada elemento de esta sucesión básica tiene un número infinito de coordenadas nulas. iv Abstract This Ph.D. dissertation mainly focuses on three multilinear problems and its aim is to describe analytical and topological techniques that we found useful to tackle these problems. The first problem comes from Quantum Information theory, it is the so-called the Separability Problem, and the other two were proposed by Gurariy. Let Mk denote the set of complex matrices of order k and let Pk be the set of positive semidefinite Hermitian matrices of Mk. The aim of this problem is to find a deterministic criterion to distinguish the separable states from the entangled states. In this work we shall only deal with the bipartite finite dimensional case, therefore the states are elements in the tensor product space Mk ⊗Mm. We say that B ∈ Mk ⊗Mm is separable if B = ∑i n =1 Ci ⊗Di, where Ci ∈ Pk and Di ∈ Pm, for every i. If B is not separable then B is entangled. Denote by VMkV the set {V XV,X ∈ Mk}, where V ∈ Mk is an orthogonal projection. We say that a linear transformation T : VMkV →WMmW is a positive map, if T (Pk ∩VMkV ) ⊂ Pm ∩WMmW . We say that a non null positive map T : VMkV → VMkV is irreducible if V ′ MkV ′ ⊂ VMkV is such that T (V ′ MkV ′) ⊂ V ′ MkV ′ then V ′ = V or V ′ = 0. Let us say that T : VMkV → VMkV is a completely reducible map, if it is a positive map and if there are orthogonal projections V1, . . . , Vs ∈ Mk such that ViVj = 0 (i ≠ j), ViV = Vi (1 ≤ i ≤ s), VMkV = V1MkV1 ⊕ . . . ⊕ VsMkVs ⊕ R, R ⊥ V1MkV1 ⊕ . . . ⊕ VsMkVs satisfying: T (ViMkVi) ⊂ ViMkVi (1 ≤ i ≤ s), T � is irreducible (1 ≤ i ≤ s), T �R ≡ 0.ViMk Vi Let A = ∑n i=1 Ai ⊗ Bi ∈ Mk ⊗Mm. Define GA : Mk → Mm, as GA(X) = ∑n i=1 tr(AiX)Bi and FA : Mm → Mk, as FA(X) = ∑i n =1 tr(BiX)Ai. Our main results are the following: If A ∈ Mk ⊗Mm is positive under partial transposition (PPT) or symmetric with positive coefficients (SPC) or invariant under realignment then FA ○ GA : Mk → Mk is completely reducible. We can reduce the Separability Problem to the weakly irreducible PPT case using this theorem. Another application of our main results is the following one: If FA ○GA :Mk →Mk v is completely reducible with the only eigenvalues 1 or 0 then A is separable. Using this theorem for a matrix invariant under realignment, we can provide a different proof of the following result published recently: If there are k mutually unbiased bases in Ck then there exists another orthonormal basis which is mutually unbiased with these k bases. Hence, if Ck contains k mutually unbiased bases then Ck contains k +1. The real case follows analogously: If R2k contains k mutually unbiased bases then R2k contains k + 1. Next, our second problem was originally proposed by Gurariy and, later, studied by Gurariy and Quarta. Let K be a topological space. Consider C(K) the vector space of real­ valued continuous functions with domain K. Denote by C�(K) the subset of C(K) formed by those functions that attain their maximum at only one point of K. The set C�(K) fails to be a vector space for many reasons, for example the zero function is not an element of this set. Gurariy and Quarta asked the following question: Can we find a subspace V of C(K) inside C�(K) ∪ {0}? If so, how big can be the dimension of V ? We have proved that if K is a compact subset of Rn and if V is a subspace of C(K) inside C�(K) ∪ {0} then dim(V ) ≤ n. This result follows from Borsuk-Ulam theorem. Our third and final problem was proposed by Gurariy and Aron. Is it possible to obtain an infinite dimensional closed subspace of �∞ such that every sequence of this space has finitely many zero coordinates? If X stands for c0 or �p, p ∈ [1, +∞], we shall denote by Z(X) the subset of X formed by sequences having only a finite number of zero coordinates. Here, we provide the definitive answer to this question: (i) There is no infinite dimensional closed subspace of X inside Z(X) ∪ {0}. (ii) There exists an infinite dimensional closed subspace of X inside V ∖ Z(V ) ∪ {0}, for any infinite dimensional closed subspace V of X. In order to obtain these results, we construct basic sequences within any infinite dimen­ sional closed subspace of X, satisfying special properties. One of this properties is each element of this basic sequence has infinitely many zero coordinates. vi Introducción Esta tesis doctoral se centra principalmente en tres problemas multilineales y su objetivo es describir las técnicas analíticas y topológicas útiles para “atacar” estos problemas. El primer problema tiene su origen en la Teoría de Información Cuántica, es el llamado problema de la separabilidad de los estados cuánticos, y los otros dos fueron propuestos por Vladimir I. Gu­ rariy. A grandes rasgos, en nuestro primer problema se emplea la teoría de Perron-Frobenius, que está relacionada con las aplicaciones positivas que actúan sobre C ∗− álgebras, con el fin de obtener una reducción del problema de la separabilidad a un caso particular y algunas otras aplicaciones a la teoría de información cuántica. Para el segundo problema, se utilizó el teorema de Borsuk-Ulam para demostrar que la dimensión de cierto espacio vectorial debe estar comprendida en cierto intervalo. Para el tercer problema, hemos construido sucesiones básicas con propiedades especiales a fin de obtener una solución completa del mismo. Denotemos por Mk al conjunto de las matrices complejas de orden k y Pk será el conjunto de matrices Hermíticas semidefinidas positivas de Mk. El problema de la separabilidad de los estados cuánticos es un problema famoso y bien establecido en el campo de la teoría de información cuántica debido a su importancia y, sobre todo, a su gran dificultad. El objetivo de este problema es encontrar un criterio determinístico para distinguir los estados separables de los estados entrelazados. Aqui sólo trabajamos con el caso bipartito de dimensión finita, luego los estados son los elementos del producto tensorial Mk ⊗Mm, que pueden ser interpretados como matrices en Mkm a través del producto de Kronecker. Decimos que B ∈ Mk ⊗Mm es separable si B = ∑i n =1 Ci ⊗Di, donde Ci ∈ Pk y Di ∈ Pm, para cada i. Si B no es separable entonces B está entrelazada. Este problema fue resuelto por completo por Horodecki en el espacio Mk ⊗Mm donde km ≤ 6, por el llamado criterio PPT (ver [29]). Este criterio establece que una matriz A = ∑k i=1 Ai ⊗Bi ∈ Mk ⊗Mm ≃ Mkm, km ≤ 6, es separable si y sólo si A permanece positiva bajo transposición parcial (PPT), es decir, A y At2 = ∑i k =1 Ai ⊗Bi t son matrices Hermíticas semidefinidas positivas (definición 3.1). 1 El caso general, incluso para el caso de dimensión finita, sigue siendo un gran desafío. Se han desarrollado algoritmos con el fin de resolver el problema de la separabilidad, pero se sabe que este problema es NP-hard (véase [28]). Por lo tanto, cualquier restricción del problema a un conjunto más pequeño de matrices es, sin duda, muy importante. Por ejemplo, Peres en [39] fue el primero en darse cuenta de la importancia de la propiedad PPT que más tarde se demostró ser necesaria y suficiente para la separabilidad en Mk ⊗Mm de km ≤ 6, en [29] . Otra reducción ha sido obtenida para el caso positivo definido en Mk ⊗Mm. Con el fin de encontrar las matrices Hermíticas positivas definidas y separables sólo tenemos que distinguir las matrices separables entre las matrices positivas definidas del tipo siguiente: l Id ⊗ Id +∑aiEi ⊗ Fi i=1 donde tr(Ei) = tr(Fi) = 0, {E1, ..., El}, {F1, ..., Fl} son conjuntos de matrices ortonormales hermíticas con respecto al producto interno de la traza y ai ∈ R. Este resultado se obtiene por medio de una forma normal (véase la subsección 3.3.2 y [23,34,46]). Los autores de [34] también obteneran una notable reducción del problema de separabili­ dad en M2 ⊗M2 para el caso general, no sólo para el caso positivo definido. Ellos mostraron que, para resolver el problema en M2 ⊗M2, es suficiente descubrir qué matrices de la siguiente familia son separables: Id ⊗ Id + d2γ2 ⊗ γ2 + d3γ3 ⊗ γ3 + d4γ4 ⊗ γ4, donde d2, d3, d4 ∈ R y γ2, γ3, γ4 son las matrices de Pauli diferentes de Id. Ellos demostraron que una matriz de esta familia es separable si y sólo si es PPT, y si y sólo si �d2�+ �d3�+ �d4� ≤ 1. Esta es una segunda demonstración del criterio PPT en M2 ⊗M2. El lector interesado puede encontrar más información en relación con el problema de separabilidad en [25]. A continuación, vamos a describir cómo utilizamos la teoría Perron-Frobenius con el fin de reducir el problema de separabilidad a un cierto subconjunto de matrices PPT y para obtener algunas otras aplicaciones. Denotemos por VMkW el conjunto {V XW,X ∈ Mk}, donde V, W ∈ Mk son proyecciones ortogonales. Si V = W entonces el conjunto VMkV es una C ∗− subálgebra hereditaria de Mk. Se dice que una transformación lineal T : VMkV → WMmW es una aplicación positiva, si T (Pk ∩ VMkV ) ⊂ Pm ∩WMmW . Se dice que una aplicación no nula positiva T : VMkV → VMkV es irreducible si V ′ MkV ′ ⊂ VMkV es tal que T (V ′ MkV ′) ⊂ V ′ MkV ′ entonces V ′ = V o V ′ = 0. Por la teoría de Perron-Frobenius, sabemos que si T : VMkV → VMkV es una aplicación positiva, entonces su radio espectral, λ, es un valor propio y hay 0 ≠ γ ∈ Pk ∩ VMkV de tal 2 manera que T (γ) = λγ. Por otra parte, si T : VMkV → VMkV es irreducible, la multiplicidad del radio espectral es 1 y las imágenes de γ y V son iguales (ver proposiciones 2.3 y 2.5 en [21]). Para ciertos tipos de aplicaciones positivas vale el recíproco del último teorema. Por ejemplo, si T : VMkV → VMkV es una aplicación positiva autoadjunta con respecto al producto interno de la traza (⟨X, Y ⟩ = tr(XY ∗)), si su radio espectral tiene multiplicidad 1 y I(γ) = I(V ) entonces T : VMkV → VMkV es irreducible (ver lema 2.11). Otro ejemplo es una aplicación completamente positiva (véase la definición en [44]). Una extensión natural del concepto de aplicación irreducible positiva es una suma directa de aplicaciones irreducibles positivas. Digamos que T : VMkV → VMkV es una aplicación completamente reducible, si es positiva y si hay proyecciones ortogonales V1, . . . , Vs ∈ Mk tales que ViVj = 0 (i ≠ j), ViV = Vi (1 ≤ i ≤ s), VMkV = V1MkV1 ⊕ . . . ⊕ VsMkVs ⊕ R, R ⊥ V1MkV1 ⊕ . . . ⊕ VsMkVs y que satisfacen: T (ViMkVi) ⊂ ViMkVi (1 ≤ i ≤ s), T � esViMkVi irreducible (1 ≤ i ≤ s), T �R ≡ 0. Observe que cualquier aplicación irreducible es completamente reducible. Este concepto está relacionado con el de la matriz completamente reducible (ver [42]). La única restricción fuerte en la definición de aplicación completamente reducible es T �R ≡ 0. Por ejemplo, la existencia de las subálgebras ViMkVi que cumplen las condiciones requeridas está garantizada para cualquier aplicación positiva autoadjunta, sin embargo la condición T �R ≡ 0 es (en general) falsa. La aplicación positiva autoadjunta más simple que no es completamente reducible es la identidad Id : Mk →Mk, k > 1. Nuevamente, como ocurre con las aplicaciones irreducibles, para aplicaciones autoadjuntas hay una propiedad simple equivalente a la propiedad de ser completamente reducible (proposición 2.13). Llamamos a esta propiedad de propiedad de descomposición (definición 2.10). Ahora vamos a centrarnos en determinados tipos de aplicaciones autoadjuntas positivas. Sea A = ∑i n =1 Ai ⊗ Bi ∈ Mk ⊗Mm e identifique Mk ⊗Mm ≃ Mkm, a través del producto de Kronecker. Defina GA : Mk → Mm, GA(X) = ∑n i=1 tr(AiX)Bi y FA : Mm → Mk, FA(X) = ∑n i=1 tr(BiX)Ai. Si A ∈ Mk ⊗Mm ≃ Mkm es Hermítica entonces FA y GA son adjuntas con respecto al producto interno de la traza. Por otra parte, si A ∈ Pkm entonces FA y GA son positivas y FA ○GA :Mk →Mk es una aplicación positiva autoadjunta. Sea S4 el grupo de permutaciones de {1, 2, 3, 4} y considere la notación de ciclos. Sea σ ∈ S4 y defina Lσ : Mk ⊗ Mk → Mk ⊗ Mk como la transformación lineal que satisface t t t tLσ(v1v2 ⊗ v3v4) = vσ(1)vσ(2) ⊗ vσ(3) vσ(4), para todos v1, v2, v3, v4 ∈ Ck. Defina Pσ = {A ∈ Mk ⊗Mk, A ∈ Pk2 and Lσ(A) ∈ Pk2 } y Iσ = {A ∈ Mk ⊗Mk, A ∈ Pk2 and Lσ(A) = A}. Entre estes tipos de matrices estamos interesados especialmente en 3: (1) P(34), que es el conjunto de las matrices PPT (definición 3.1) (2) P(243), que es el conjunto de las matrices SPC (definiton 3.6) 3 (3) I(23), que es el conjunto de las matrices invariantes por realineamiento (definición 3.8). Finalmente, podemos describir nuestros principales resultados. Si A ∈ Mk ⊗Mm es positiva bajo transposición parcial (PPT) o simétrica con coeficientes positivos (SPC) o invariante bajo realineamiento luego FA ○GA :Mk →Mk es completamente reducible (teoremas 3.2, 3.12 y 3.13). Vamos a aplicar nuestros principales resultados a la teoría de información cuántica. La aplicación FA ○GA :Mk →Mk es responsable por la descomposición de Schmidt de la matriz Hermítica A ∈ Mk ⊗Mm. Nuestros principales teoremas dicen que bajo una de estas tres hipótesis la aplicación FA ○GA :Mk →Mk se descompone como una suma de aplicaciones irreducibles. Por lo tanto, A también se descompone como una suma de matrices débilmente irreducibles (definición 3.15 y la proposición 3.18). Una condición necesaria para la separabilidad de A ∈ Mk ⊗Mm es ser PPT. Podemos utilizar la descomposición de una matriz PPT como una suma de matrices débilmente irre­ ducibles para reducir el problema de separabilidad para el caso PPT débilmente irreducible (corolario 3.20). También proporcionamos una descripción completa de las matrices PPT débilmente irreducibles (proposición 3.17). Una herramienta importante para estudiar la separabilidad de las matrices Hermitícas definidas positivas en Mk ⊗ Mm es una forma normal denominada forma nomal de filtro (véase la sección IV.D de [23] y la subsección 3.3.2). La única prueba conocida de esta forma normal depende de A ser positiva definida. En realidad, la descomposición de una matriz PPT como una suma de matrices débilmente irreducibles proporciona otro caso en el que la forma normal de filtro puede ser utilizada (véase la subsección 3.3.2). Esto plantea una pregunta importante: ¿Podemos demostrar la forma normal de filtro para las matrices PPT débilmente irreducibles? Si la respuesta fuera sí, seríamos capazes de usar la forma normal de filtro para cualquier matriz PPT. Todavía podemos obtener algunas desigualdades que impliquen separabilidad para matri­ ces PPT débilmente irreducibles, incluso sin la forma normal de filtro. Estas desigualdades se basan en el hecho de que toda matriz Hermítica positiva semidefinida con rango tensorial 2 es separable (ver teorema 3.44). Queremos enfatizar que la forma normal de filtro también sería útil para mejorar estas desigualdades (ver ejemplo 3.38). Otra de las aplicaciones de nuestros resultados principales es la siguiente: Si FA ○GA : Mk →Mk es completamente reducible con los únicos valores propios 1 o 0 entonces A es sep­ arable. El uso de este teorema para una matriz invariante bajo realineamiento, proporciona una prueba diferente del siguiente resultado publicado recientemente en [47]: Si hay k bases mutuamente imparciales en Ck entonces existe otra base ortonormal que es mutuamente imparcial con estas k bases. Por lo tanto, si Ck contiene k bases mutuamente imparciales entonces Ck contiene k + 1. El caso real sigue de manera análoga: Si R2k contiene k bases mutuamente imparciales entonces R2k contiene k + 1. 4 Este resultado es bastante sorprendente, ya que algunos conjuntos de bases mutuamente imparciales se demostraron inextensibles (véase, por ejemplo, [36]). En la teoría de infor­ mación cuántica, el concepto de bases mutuamente imparciales (definición 3.23) se ha de­ mostrado útil. Tiene aplicaciones en tomografía y criptografía (vea [17, 31, 48, 49]) . Se sabe que k + 1 es un límite superior para el número de bases mutuamente imparciales de Ck y la existencia de este número de bases es un problema abierto, cuando k no es una potencia de números primos. Cuando k es una potencia de cierto número primo, se han utilizados métodos constructivos para obtener estos k + 1 bases (ver [4, 31,49]). Además de la información que nuestros principales teoremas proporcionan, también nos proporcionan una intuición: Los tres tipos de matrices que aparecen en nuestros teoremas están conectados. Por lo tanto, podemos preguntarnos si cada matriz SPC es PPT o si cada matriz invariante bajo realineamiento es PPT. Se demuestra que las matrices SPC y las matrices invariantes bajo realineamiento son PPT en M2 ⊗M2, sin embargo, en Mk ⊗Mk, k > 2, hay contraejemplos. Observemos que la propiedad de ser completamente reducible es muy fuerte. Es una sorpresa que FA ○GA : Mk → Mk sea completamente reducible, cuando A es PPT o SPC o invariante bajo realineamiento. Una matriz PPT es un tipo muy común de estado en teoría de información cuántica. Por otra parte, se sabe que un estado invariante bajo la multiplicación por el operador "Flip" es PPT si y sólo si es SPC (ver [45] y la proposición 3.33), por lo tanto, las matrices SPC son relativamente conocidas. Matrices invariante bajo realineamiento no son muy comunes, pero el realineamiento es bien conocido debido a su uso con el fin de detectar entrelazamiento. Muy a menudo la teoría de información cuántica se beneficia de las ideas y de los teoremas de la teoría de las aplicaciones positivas. Por ejemplo, la solución completa del problema de la separabilidad en Mk ⊗Mm, km ≤ 6, se obtuvo mediante la clasificación completa de las aplicaciones positivas, T :Mk →Mm, km ≤ 6. Como la teoría de información cuántica está proporcionando sus tipos especiales de estados como la hipótesis de nuestros principales resultados, podemos interpretar estos resultados como la retroalimentación de esta teoría a la teoría de aplicaciones positivas. Nuestro segundo problema fue propuesto originalmente por Vladimir I. Gurariy y, más tarde, estudiado por Gurariy y Quarta en [27]. Sea K un espacio topológico. Consideremos C(K) el espacio vectorial de las funciones reales continuas con dominio K. Denotemos por C�(K) el subconjunto de C(K) formado por aquellas funciones que alcanzan su máximo en un solo punto de K. El conjunto C�(K) no es un espacio vectorial por muchas razones, por ejemplo, la función cero no es un elemento de este conjunto. Gurariy y Quarta se hicieron la siguiente pregunta: ¿Podemos encontrar un subespacio V de C(K) dentro de C�(K) ∪ {0}? En caso positivo, ¿cómo de grande puede llegar a ser la dimensión de V ? Los principales resultados obtenidos por Gurariy y Quarta en este sentido son los sigu­ ientes: 5 (A) Existe un subespacio de dimensión 2 de C[a, b) contenido en C�[a, b) ∪ {0}. (B) Hay un subespacio de dimensión 2 de C(R) contenido en C�(R) ∪ {0} . (C) No existe ningún subespacio de dimensión 2 de C [a, b] contenido en C� [a, b] ∪ {0}. Nuestro principal resultado es una generalización de (C). Hemos demostrado que si K es un subconjunto compacto de Rn y si V es un subespacio de C(K) dentro C�(K) ∪ {0} entonces dim(V ) ≤ n. Mientras que Gurariy y Quarta [27] emplearon técnicas analíticas clásicas, nuestra generalización requiere de un teorema topológico: el famoso teorema de Borsuk-Ulam. La razón por la cual el teorema de Borsuk-Ulam es útil en este contexto es la siguiente: Supongamos que f1, . . . , fk es una base de un subespacio V de C(K) dentro C�(K)∪{0}. Sea Sk−1 la esfera Euclídea dentro de Rk y definamos la función g : Sk−1 →K como g(a1, . . . , ak) = el único punto de máximo en K de ∑k i=1 aifi. Demostramos que esta función es continua si K es un subconjunto compacto de Rn. Por el teorema de Borsuk-Ulam, si la dimensión k del subespacio V es mayor que n, entonces hay un par de puntos antipodales en Sk−1 con la misma imagen. Por lo tanto, tenemos que hay f, −f dentro de este subespacio con el mismo punto de máximo. Entonces, f es constante y no alcanza su máximo en un solo punto, lo cual es absurdo. En general, la función g no es continua. Por ejemplo, si K = [0, 2π) entonces el subespacio de C([0, 2π)) generado {cos(t), sin(t)} es un subconjunto de C�([0, 2π)) ∪ {0}. La función g : S1 → [0, 2π), g(a1, a2) = el único punto de máximo en [0, 2π) de a1 cos(t) + a2 sin(t), no es continua en (1, 0). La continuidad de g bajo la hipótesis de compacidad de K es una sorpresa. Nuestro tercer y último problema fue propuesto por Richard M. Aron Y Vladimir I. Gurariy. ¿Es posible obtener un subespacio cerrado de dimensión infinita de �∞ tal que cada sucesión de este espacio tiene solamente un número finito de coordenadas nulas? Esta cuestión ha aparecido en varios trabajos recientes (véase, por ejemplo, [9, 20, 22, 38]) y, durante la última década, ha habido varios intentos de responder parcialmente, aunque no hay nada concluyente en relación con el original problema que se ha obtenido hasta ahora. Si X denota un espacio de sucesiones, designaremos por Z(X) el subconjunto de X formado por sucesiones que tienen sólo un número finito de coordenadas cero. A continuación, mostraremos (entre otros resultados) la respuesta definitiva a esta pregunta. Es decir, si X representa c0 o �p, con p ∈ [1, ∞], se prueba lo siguiente: 6 (i) No hay subespacio cerrado de dimensión infinita de X dentro Z(X) ∪ {0} (corolarios 5.7 y 5.16). (ii) Hay un subespacio cerrado de dimensión infinita de X dentro V ∖Z(V ), para cualquier subespacio cerrado V (con dimensión infinita) de X (teorema 5.18). Para obtener estos resultados, se construyen sucesiones básicas dentro de cualquier sube­ spacio cerrado de dimensión infinita de X = c0 o �p, p ∈ [1, +∞], y verificando propiedades “especiales”. Una de estas propiedades es que cada elemento de esta sucesión básica tiene un número infinito de coordenadas nulas. Observemos que hay un ejemplo muy simple de un subespacio de dimensión infinita de X dentro Z(X) ∪ {0}, que por supuesto, no es cerrado. Sea V el subespacio generado por {(λn)n∈N � 0 < λ < 1}. Tenga en cuenta que V ⊂ X, si X = c0 o �p, p ∈ [1, +∞], y cualquier combinación lineal no trivial de distintos (λn)n∈N, . . . , (λn)n∈N está dominado por (λn)n∈N con1 k i el mayor λi. Por lo tanto, las coordenadas de esta combinación lineal no son cero después de una cierta coordenada, que depende de la combinación. Por lo tanto, V ⊂ Z(X) ∪ {0}. Usando terminología moderna (acuñada originalmente por V.I. Gurariy), un subconjunto M de un espacio vectorial topológico X se llama lineable (resp. espaciable) en X si existe un subespacio lineal de dimensión infinita (resp. subespacio lineal cerrado de dimensión infinita) S ⊂ M ∪{0} (véase [1,2,5,9,10,20,27]). Por lo tanto, hemos demostrado que Z(X) es lineable y no espaciable. No hay muchos ejemplos de conjuntos (no triviales) que son lineables y no espaciables. Uno de los primeros en este sentido, se debe a Levine y Milman (1940, [35]) que mostraran que el subconjunto de C[0, 1] de todas las funciones de variación acotada no es espaciable (obviamente lineable, ya que es un espacio lineal de dimensión infinita). Una más reciente se debe a Gurariy (1966, [26]), que probó que el conjunto de funciones diferenciables en todas partes [0, 1] (que es también un espacio lineal de dimensión infinita) no es espaciable en C([0, 1]). Sin embargo, Bernal-González ([8], 2010) mostró que C∞(]0, 1[) es, en realidad, espaciable en C(]0, 1[). 7 8 Chapter 1 Introduction This Ph.D. dissertation mainly focuses on three multilinear problems and its aim is to describe analytical and topological techniques that we found useful to tackle these problems. The first problem comes from Quantum Information theory, it is the so-called the Separability Problem, and the other two were proposed by Gurariy. In our first problem we used Perron- Frobenius Theory, which is related to positive maps acting on C ∗−algebras, in order to obtain a reduction of the Separability Problem to a particular case and some other applications to Quantum Information theory. For the second problem, we used Borsuk-Ulam theorem to show that the dimension of a particular vector space must be within a certain range in order to exist. For the third problem, we constructed basic sequences with special properties in order to obtain a complete solution. Let Mk denote the set of complex matrices of order k and let Pk be the set of positive semidefinite Hermitian matrices of Mk. The Separability Problem is a well established prob­ lem in the field of Quantum Information Theory due to its importance and difficulty. The aim of this problem is to find a deterministic criterion to distinguish the separable states from the entangled states. In this work we shall only deal with the bipartite finite dimensional case, therefore the states are elements in the tensor product space Mk ⊗Mm, which can be interpreted as matrices in Mkm via the Kronecker product. We say that B ∈ Mk ⊗Mm is separable if B = ∑i n =1 Ci ⊗Di, where Ci ∈ Pk and Di ∈ Pm, for every i. If B is not separable then B is entangled. This problem was completely solved by Horodecki in the space Mk ⊗Mm for km ≤ 6, by the so-called PPT criterion (see [29]). This criterion states that a matrix A = ∑i n =1 Ai ⊗Bi ∈ Mk ⊗Mm ≃ Mkm, km ≤ 6, is separable if and only if A is positive under partial transposition (PPT), i.e., A and At2 = ∑i n =1 Ai ⊗Bi t are positive semidefinite Hermitian matrices (definition 3.1). 9 The general case, even for the finite dimensional case, is still a great challenge. Algorithms have been developed in order to solve the separability problem, but it is known that this problem is NP-hard (see [28]). Therefore, any restriction of the problem to a smaller set of matrices is, certainly, important. For example, Peres in [39] was the first to notice the importance of the PPT property which was later proved to be necessary and sufficient for separability in Mk ⊗Mm for km ≤ 6, in [29]. Another remarkable reduction was obtained for the positive definite case in Mk ⊗Mm. In order to find the separable positive definite Hermitian matrices we only need to distinguish the separable matrices among the positive definite matrices of the following type: l Id ⊗ Id +∑aiEi ⊗ Fi i=1 where tr(Ei) = tr(Fi) = 0, {E1, ..., El}, {F1, ..., Fl} are orthonomal sets of Hermitian matrices with respect to the trace inner product and ai ∈ R. This result is obtained via the filter normal form (see subsection 3.3.2 and [23,34,46]). The authors of [34] also obtained a remarkable reduction of the separability problem in M2 ⊗M2 for the general case, not only for the positive definite case. They showed that, in order to solved it, it suffices to discover which matrices from the following family of matrices are separable: Id ⊗ Id + d2γ2 ⊗ γ2 + d3γ3 ⊗ γ3 + d4γ4 ⊗ γ4, where d2, d3, d4 ∈ R and γ2, γ3, γ4 are the matrices of the Pauli’s basis of M2 different from the Id. They proved that a matrix within this family is separable if and only if it is PPT, and if and only if �d2�+ �d3�+ �d4� ≤ 1. This is a second proof of the PPT criterion in M2 ⊗M2. The interested reader can find more information concerning the Separability Problem in the survey [25]. Next, let us describe how we used the Perron-Frobenius theory in order to reduce the sep­ arability problem to a certain subset of PPT matrices and to obtain some other applications. Denote by VMkW the set {V XW,X ∈ Mk}, where V, W ∈ Mk are orthogonal projections. If V = W then the set VMkV is an hereditary C ∗−subalgebra of Mk. We say that a linear transformation T : VMkV →WMmW is a positive map, if T (Pk ∩ VMkV ) ⊂ Pm ∩WMmW . We say that a non null positive map T : VMkV → VMkV is irreducible if V ′ MkV ′ ⊂ VMkV is such that T (V ′ MkV ′) ⊂ V ′ MkV ′ then V ′ = V or V ′ = 0. By Perron-Frobenius theory, we know that if T : VMkV → VMkV is a positive map then its spectral radius, λ, is an eigenvalue and there is 0 ≠ γ ∈ Pk ∩ VMkV such that T (γ) = λγ. Moreover, if T : VMkV → VMkV is irreducible then the multiplicity of the spectral radius is 1 and the images of γ and V are equal (see propositions 2.3 and 2.5 in [21]). 10 There are certain types of positive maps such that the converse of the last theorem is valid. For example, if T : VMkV → VMkV is a self-adjoint positive map with respect to the trace inner product (⟨X, Y ⟩ = tr(XY ∗)), if its spectral radius has multiplicity 1 and I(γ) = I(V ) then T : VMkV → VMkV is irreducible (see lemma 2.11). Another example is a completely positive map (see definition 1 in [44]) . A natural extension of the concept of irreducible positive map is a direct sum of irreducible positive maps. Let us say that T : VMkV → VMkV is a completely reducible map, if it is a positive map and if there are orthogonal projections V1, . . . , Vs ∈ Mk such that ViVj = 0 (i ≠ j), ViV = Vi (1 ≤ i ≤ s), VMkV = V1MkV1 ⊕. . .⊕VsMkVs ⊕R, R ⊥ V1MkV1 ⊕. . .⊕VsMkVs satisfying: T (ViMkVi) ⊂ ViMkVi (1 ≤ i ≤ s), T �ViMkVi is irreducible (1 ≤ i ≤ s), T �R ≡ 0. Notice that any irreducible map is completely reducible. This concept is related to that of completely reducible matrix (see [42]). The only strong restriction in the definition of completely reducible map is T �R ≡ 0. For example, the existence of the subalgebras ViMkVi satisfying the required conditions is granted for any self-adjoint positive map, however the condition T �R ≡ 0 is (in general) false. The simplest self-adjoint positive map that is not completely reducible is the identity map Id : Mk → Mk, k > 1. As for irreducible maps, if T : VMkV → VMkV is a self-adjoint map then there is a very neat property equivalent to the complete reducibility of T (proposition 2.13). We call this property the decomposition property (definition 2.10). Now, let us focus on specific types of self-adjoint positive maps. Let A = ∑i n =1 Ai ⊗Bi ∈ Mk ⊗Mm and identify Mk ⊗Mm ≃ Mkm, via Kronecker product. Define GA : Mk → Mm, as GA(X) = ∑i n =1 tr(AiX)Bi and FA :Mm →Mk, as FA(X) = ∑i n =1 tr(BiX)Ai. If A ∈ Mk ⊗Mm ≃ Mkm is Hermitian then FA and GA are adjoint with respect to the trace inner product. Moreover, if A ∈ Pkm then FA and GA are positive maps and FA ○ GA : Mk → Mk is a self-adjoint positive map. Next, let S4 be the group of permutations of {1, 2, 3, 4} and consider the cycle notation. Let σ ∈ S4 and define Lσ : Mk ⊗Mk → Mk ⊗Mk as the linear transformation that satisfies t t t tLσ(v1v2 ⊗ v3v4) = vσ(1)vσ(2) ⊗ vσ (3) v σ(4), for every v1, v2, v3, v4 ∈ Ck. Define Pσ = {A ∈ Mk ⊗ Mk, A ∈ Pk2 and Lσ(A) ∈ Pk2 } and Iσ = {A ∈ Mk ⊗Mk, A ∈ Pk2 and Lσ(A) = A}. Among these types of matrices we are specially interested in 3 types: (1) P(34), which is the set of PPT matrices (definition 3.1) (2) P(243), which is the set of SPC matrices (definiton 3.6) (3) I(23), which is the set of matrices invariant under realignment (definiton 3.8). We can finally describe our main results. If A ∈ Mk ⊗Mm is positive under partial trans­ position (PPT) or symmetric with positive coefficients (SPC) or invariant under realignment 11 then FA ○ GA : Mk → Mk is completely reducible (theorems 3.2, 3.12 and 3.13). We shall apply our main results to Quantum Information Theory. The map FA○GA :Mk →Mk is responsible for the Schmidt decomposition of the Hermitian matrix A. Our main theorems say that under one of these three hypothesis the map FA ○GA : Mk → Mk decomposes as a sum of irreducible maps. Hence, A shall also decomposes as a sum of weakly irreducible matrices (definition 3.15 and proposition 3.18). A necessary condition for the separability of A ∈ Mk ⊗Mm is to be PPT. We can use the decomposition of a PPT matrix as a sum of weakly irreducible matrices to reduce the Separability Problem to the weakly irreducible PPT case (corollary 3.20). We also provide a complete description of weakly irreducible PPT matrices (proposition 3.17). An important tool to study separability of positive definite Hermitian matrices in Mk ⊗Mm is the so-called filter normal form (see section IV.D of [23] and subsection 3.3.2). The only known proof of this normal form depends heavily on the positive definiteness of A. Actually, the decomposition of a PPT matrix as a sum of weakly irreducible matrices provides another case where the filter normal form can be used (see subsection 3.3.2). This raises an important question: Can we prove the filter normal form for weakly irreducible PPT matrices? If so, we would be able to use the filter normal form for every PPT matrix. We can still obtain some inequalities for weakly irreducible PPT matrices that imply separability, even without the filter normal form. These inequalities are based on the fact that every positive semidefinite Hermitian matrix with tensor rank 2 is separable (see theorem 3.44). We want to emphasize that the filter normal form would also be useful to sharpen these inequalities (see example 3.38). Another application of our main results is the following one: If FA ○ GA : Mk → Mk is completely reducible with the only eigenvalues 1 or 0 then A is separable. Using this theorem for a matrix invariant under realignment, we can provide a different proof of the following result published recently in [47]: If there are k mutually unbiased bases in Ck then there exists another orthonormal basis which is mutually unbiased with these k bases. Hence, if Ck contains k mutually unbiased bases then Ck contains k + 1. The real case follows analogously: If R2k contains k mutually unbiased bases then R2k contains k + 1. This result is quite surprising, since some sets of mutually unbiased bases were proved to be unextendible (see, e.g., [36]). In Quantum Information Theory, the concept of mutually unbiased bases (definition 3.23) has been shown to be useful. It has applications in state determination, quantum state tomography, cryptography (see [17, 31, 48, 49]). It is known that k + 1 is an upper bound for the number of mutually unbiased bases in Ck and the existence of this number of bases is an open problem, when k is not the power of prime number. When k is a power of certain prime number, some constructive methods were used to obtain these k + 1 bases (see [4, 31, 49]). 12 Besides the information that our main theorems provide, they also provide an intuition: The three types of matrices that our main theorems concern are connected. Thus, we can wonder if every SPC matrix is PPT or if every matrix invariant under realignment is PPT. We show that SPC matrices and matrices invariant under realignment are PPT in M2 ⊗M2, however in Mk ⊗Mk, k > 2, there are counterexamples. Notice that the complete reducibility of FA ○GA : Mk →Mk is a very stong property. It is quite a surprise that FA ○GA : Mk → Mk is completely reducible, if A is PPT or SPC or invariant under realignment. A PPT matrix is a very common type of state in Quantum Information Theory. Moreover, it is known that a state invariant under multiplication by the flip operator is PPT if and only if is SPC (see [45] and proposition 3.33), thus SPC matrices are relatively known. Matrices invariant under realignment are not very common, but the realignment map is well known due to its use in order to detect entanglement. Very often Quantum Information Theory benefits from the ideas and theorems of the theory of positive maps. For example the complete solution of the Separability Problem in Mk ⊗Mm, km ≤ 6, was obtained by the complete classification of the positive maps, T : Mk → Mm, km ≤ 6. Since Quantum Information Theory is providing its special types of states as the hypothesis of our main results, we can interpret these results as the feedback of this theory to the theory of positive maps. Next, our second problem was originally proposed by Gurariy and, later, studied by Gurariy and Quarta in [27]. Let K be a topological space. Consider C(K) the vector space of real-valued continuous functions with domain K. Denote by C�(K) the subset of C(K) formed by those functions that attain their maximum at only one point of K. The set C�(K) fails to be a vector space for many reasons, for example the zero function is not an element of this set. Gurariy and Quarta asked the following question: Can we find a subspace V of C(K) inside C�(K) ∪ {0}. If so, how big can be the dimension of V ? The main results obtained by Gurariy and Quarta in this direction are the following: (A) There is a 2-dimensional linear subspace of C[a, b) contained in C�[a, b) ∪ {0}. (B) There is a 2-dimensional linear subspace of C(R) contained in C�(R) ∪ {0}. (C) There is no 2-dimensional linear subspace of C [a, b] contained in C� [a, b] ∪ {0}. Our main result is a generalization of (C). We proved that if K is a compact subset of Rn and if V is a subspace of C(K) inside C�(K) ∪ {0} then dim(V ) ≤ n. While Gurariy and Quarta [27] used typical analytical techniques, our generalization requires a topological theorem: Borsuk-Ulam theorem. 13 The reason why the Borsuk-Ulam theorem is useful within this context is the following: Assume that f1, . . . , fk is a basis of a subspace V of C(K) inside C�(K) ∪ {0}. Let Sk−1 be the Euclidean sphere inside Rk and define the function g : Sk−1 → K as g(a1, . . . , ak) = the unique point of maximum in K of ∑k i=1 aifi. We proved that this function is continuous if K is a compact subset of Rn. By Borsuk-Ulam theorem, if the dimension k of the subspace V is bigger than n then there is a pair of antipodal points in Sk−1 with the same image. Thus, there are f, −f inside this subspace with the same point of maximum. Hence, f is constant and does not attain its maximum at only one point, which is absurd. In general, the function g is not continuous. For example, if K = [0, 2π) then the subspace span{cos(t), sin(t)} of C([0, 2π)) is a subset of C�([0, 2π))∪{0}. The function g : S1 → [0, 2π), g(a1, a2) = the unique point of maximum in [0, 2π) of a1 cos(t) + a2sin(t), is not continuous at (1, 0). It was quite a surprise to obtain the continuity of g under the hypothesis of compactness of K. Our third and final problem was proposed by Gurariy and Aron. Is it possible to obtain an infinite dimensional closed subspace of �∞ such that every sequence of this space has finitely many zero coordinates? This question has appeared in several recent works (see, e.g., [9, 20, 22, 38]) and, for the last decade, there have been several attempts to partially answer it, although nothing conclusive in relation to the original problem has been obtained so far. If X denotes a sequence space, we shall denote by Z(X) the subset of X formed by sequences having only a finite number of zero coordinates. Here, we shall provide (among other results) the definitive answer to this question. Namely, if X stands for c0, or �p, with p ∈ [1, ∞], we prove the following: (i) There is no infinite dimensional closed subspace of X inside Z(X) ∪ {0} (Corollaries 5.7 and 5.16). (ii) There exists an infinite dimensional closed subspace of X inside V ∖ Z(V ) ∪ {0}, for any infinite dimensional closed subspace V of X (Theorem 5.18). In order to obtain these results, we construct basic sequences within any infinite dimen­ sional closed subspace of X = c0 or �p, p ∈ [1, +∞], satisfying special properties. One of this properties is each element of this basic sequence has infinitely many zero coordinates. Observe that there is a very simple example of an infinite dimensional subspace of X inside Z(X) ∪ {0}, of course this subspace is not closed. Consider V = span{(λn)n∈N � 0 < λ < 1}. Notice that V ⊂ X, for X = c0 or �p, p ∈ [1, +∞], and any non-trivial linear combination 14 of distinct (λn)n∈N, . . . , (λn)n∈N is dominated by (λn)n∈N with the largest λi. Hence, the 1 k i coordinates of this linear combination are not zero after a certain coordinate, which depends on the combination. Hence, V ⊂ Z(X) ∪ {0}. Using modern terminology (originally coined by Gurariy himself), a subset M of a topo­ logical vector space X is called lineable (resp. spaceable) in X if there exists an infinite dimensional linear space (resp. an infinite dimensional closed linear space) Y ⊂ M ∪ {0} (see [1, 2, 5, 9, 10,20,27]). Thus, we have proved that Z(X) is lineable and not spaceable. There are not many examples of (nontrivial) sets that are lineable and not spaceable. One of the first ones in this direction, is due to Levine and Milman (1940, [35]) who showed that the subset of C[0, 1] of all functions of bounded variation is not spaceable (it is obviously lineable, since it is an infinite dimensional linear space itself). A more recent one is due to Gurariy (1966, [26]), who showed that the set of everywhere differentiable functions on [0, 1] (which is also an infinite dimensional linear space) is not spaceable in C([0, 1]). However, Bernal-González ([8], 2010) showed that C∞(]0, 1[) is, actually, spaceable in C(]0, 1[). 15 16 Chapter 2 Some Results from the Perron-Frobenius Theory All definitions and results in this chapter can, also, be found in [14]. Let Mk denote the set of complex matrices of order k. Let us denote by VMkW the set {V XW � X ∈ Mk}, where V, W ∈ Mk are orthogonal projections. If V = W then VMkV is a hereditary finite dimensional C ∗-algebra (see [21]). Let Pk denote the set of positive semidefinite Hermitian matrices in Mk. A linear transformation L : VMkV → WMmW is said to be a positive map if L(Pk ∩ VMkV ) ⊂ Pm ∩WMmW . A non-null positive map L : VMkV → VMkV is called irreducible, if V ′ MkV ′ ⊂ VMkV is such that L(V ′ MkV ′) ⊂ V ′ MkV ′ then V ′ = V or V ′ = 0. Within the context of positive maps, sometimes the term self-adjoint means L(A ∗) = L(A)∗ (see, e.g., [21]). Here, we shall use this terminology with its usual meaning. We shall say that L : VMkV → VMkV is self-adjoint if L is equal to its adjoint L ∗ (i.e., ⟨L(A),B⟩ = ⟨A, L(B)⟩). We shall consider the usual inner product in Mk, ⟨A, B⟩ = tr(AB ∗). In this chapter, we use well known theorems from the Perron-Frobenius Theory to describe some properties of completely reducible maps (see definition 3.16). These are theorems 2.3 and 2.5 in [21]: If L : VMkV → VMkV is a positive map then there exists γ ∈ Pk ∩ VMkV such that L(γ) = λγ, where λ is the spectral radius of L. Moreover, if L is irreducible then this eigenvalue has multiplicity 1. We present elementary proofs of these theorems in Section 2.1 (see theorems 2.7 and 2.8). In Section 2.2, we prove that if L : VMkV → VMkV is a self-adjoint positive map then L is completely reducible if and only if L has the decomposition property (proposition 2.13). In the next chapter, we provide an equivalent way to prove that L has the decomposition property (lemma 3.14) and we shall give two applications of completely reducible maps to Quantum Information Theory. 17 2.1 Two theorems from the Perron-Frobenius Theory For the convenience of the reader and for the completeness of this dissertation, we present here elementary proofs of theorems 2.7 and 2.8. These theorems are well known results from the Perron-Frobenius theory (see propositions 2.3 and 2.5 in [21]). Notice that there are general versions of them in the literature (check, for example, the appendix of [43]). Here, the proofs of theorems 2.7 and 2.8 follow the ideas of [6] and [21]. The first two lemmas are well known and their proofs are omitted. Recall that a positive map L : VMkV →WMmW preserves hermiticity, since every Hermitian matrix is a difference of two positive semidefinite Hermitian matrices. Lemma 2.1. If A ∈ Pk and B ∈ Mk is Hermitian then I(B) ⊂ I(A) if and only if there exists q > 0 such that A ± qB ∈ Pk. Lemma 2.2. Let γ1, γ2 be Hermitian matrices in Mk, γ1 ∈ Pk and γ2 ≠ 0. Suppose that I(γ2) ⊂ I(γ1) and γ2 is not a multiple of γ1. There exists 0 ≠ λ ∈ R such that γ1 − λγ2 ∈ Pk and 0 ≠ v ∈ ker(γ1 − λγ2) ∩ I(γ1). Lemma 2.3. Let L : VMkV →WMmW be a positive map. If γ ∈ Pk ∩ VMkV and L(γ) = δ then L(V1MkV1) ⊂ W1MmW1, where V1 is the orthogonal projection onto I(γ) and W1 is the orthogonal projection onto I(δ). Proof. Let γ1 ∈ V1MkV1 be a Hermitian matrix. Thus, I(γ1) ⊂ I(V1) = I(γ). So there is q > 0 such that γ ± qγ1 ∈ Pk, by lemma 2.1. Now, since L(γ1) is Hermitian and L(γ) ± qL(γ1) = L(γ ± qγ1) ∈ Pk then I(L(γ1)) ⊂ I(L(γ)) = I(δ) = I(W1), by lemma 2.1. Therefore, L(γ1) ∈ W1MmW1. Finally, since every matrix in V1MkV1 is a linear combination of Hermitian matrices within V1MkV1 then L(V1MkV1) ⊂ W1MmW1. Corollary 2.4. Let L : VMkV → VMkV be a positive map and γ ∈ Pk ∩ VMkV be such that L(γ) = λγ, λ > 0. Then, L(V1MkV1) ⊂ V1MkV1, where V1 is the orthogonal projection onto I(γ). Recall that the largest absolute value of all eigenvalues of L : VMkV → VMkV is called the spectral radius of L. Lemma 2.5. Let L : VMkV → VMkV be a positive map. If L(V ) = V then the spectral radius of L is 1. 18 tProof. Let U ∈ VMkV be a normal matrix such that UU ∗ = U ∗ U = V . Thus, U = ∑i s =1 λivivi , where s is the rank of V , {λ1, . . . , λs} are complex numbers of norm 1 and {v1, . . . , vs} is an orthonormal basis of I(V ). Recall that L(U ∗) = L(U)∗, since L preserves hermiticity, and L(U ∗)V = L(U ∗), V L(U) = L(U). s L(U ∗) L(V ))( 1 λi ( L(V ) L(U)Now, consider the matrix B =∑ λi 1 )⊗L(vivit) = = i=1 V L(U) Id 0 0 L(U)= ( ) = ( )(V V −L(U)∗ L(U))(Id ) . L(U)∗ V L(U)∗ Id 0 0 Id Since B ∈ P2k then V − L(U)∗ L(U) ∈ Pk. So �L(U)�2 ≤ 1, where �L(U)�2 is the spectral norm of L(U). Thus, for every normal matrix U such that UU ∗ = U ∗ U = V , we have�L(U)�2 ≤ 1 Next, let A ∈ VMkV be an eigenvector of L associated to some eigenvalue α and �A�2 = 1. iθj −iθjLet ∑s j=1 aj mj nt j be a SVD decomposition of A, where 0 ≤ aj ≤ 1. Since aj = cos(θj ) = e + e then A = 1 (∑s j=1 e iθj mj nt )+ 1 (∑j s =1 e −iθj mj nt ) = 1 U1 + 1 U2. Notice that U1U ∗ = U ∗ U1 = 2 U2U ∗ 2 = 2 j 2 j 2 2 1 1 2 U2 ∗ U2 = V . 1 1Finally, �α� = �α��A�2 = �L(A)�2 = �L(1 U1 + U2)�2 ≤ 2 1 �L(U1)�2 + �L(U2)�2 = 1 and since 2 2 2 1 is an eigenvalue of L : VMkV → VMkV , L(V ) = V , then 1 is the spectral radius of L. Lemma 2.6. Let L : VMkV → VMkV be an irreducible positive map. If 0 ≠ X ∈ Pk ∩VMkV then I((Id +L)s−1(X)) = I(V ), where s = rank(V ). Proof. If 0 ≠ X ∈ Pk ∩ VMkV then L(X) ∈ Pk ∩ VMkV and I(x) ⊂ I(x +L(x)) ⊂ I(V ) . Now, if I(X + L(X)) = I(X) then I(L(X)) ⊂ I(X) and, by lemma 2.3, L(V ′ MkV ′) ⊂ V ′ MkV ′ , where V ′ is the orthogonal projection onto I(X). Since L is irreducible then V = V ′ and I(X) = I(V ). Thus, I(X) ≠ I(V ) implies rank((Id +L)(X)) > rank(X). Repeating the argument at most s − 1 times, we obtain I((Id +L)s−1(X)) = I(V ). Theorem 2.7. Let L : VMkV → VMkV be an irreducible positive map. The spectral radius r of L is an eigenvalue of L associated to some eigenvector Z ∈ Pk ∩VMkV such that I(Z) = I(V ). Moreover, the geometric multiplicity of r is 1. Proof. Let Z = {X ∈ Pk ∩ VMkV, I(X) = I(V )}. Define f : Z → [0, ∞[ as f(X) = sup{λ ∈ R, L(X) − λX ∈ Pk}. Denote by B+ the pseudo-inverse of B and by �B�2 the spectral norm of B. Notice that if B ∈ Z then B+B = BB+ = V and B+V = V B+ = B+. 19 Now, for every X ∈ Z, there is Y ∈ Z such that Y 2 = X. Notice that f(X) is the minimal positive eigenvalue of Y +L(X)Y +, which is the minimal positive eigenvalue of L(X)Y +Y + = L(X)X+. Next, if (An)n∈N ∈ Z converges to A ∈ Z, then the smallest positive eigenvalue of An, �A+�−1, converges to the smallest positive eigenvalue of A, �A+�−1(see pg 154 in [7]). Thus, n 2 �−1 2 there is N ∈ N such that if n > N then �A+ n 2 ≥ (2�A+�2)−1. Hence, for n > N , �A+ n�2 ≤ 2�A+�2 and �A+ n −A+�2 = �An +(A −An)A+�2 ≤ �An +�2�A+�2�An −A�2 ≤ 2�A+�22�An −A�2. Therefore A+ n converges to A+ and A+ depends continuously on A, for A varying on Z. Since the eingevalues of a matrix vary continuously with a matrix (see pg 154 in [7]) then f : Z → [0, ∞[ is a continuous function. Consider the compact set Z ′ = {X ∈ Pk ∩ VMkV, �X�2 = 1}, where �X�2 is the spectral norm of X. By lemma 2.6, (Id+L)s−1(Z ′) is a compact subset of Z. Therefore, f �(Id+L)s−1(Z ′) attains its maximum r at some point W ∈ (Id +L)s−1(Z ′) ⊂ Z. By definiton of r = f(W ), we have L(W ) − rW ∈ Pk. Next, if 0 ≠ L(W ) − rW then the range of (Id +L)s−1(L(W ) − rW ) is the range of V , by lemma 2.6. Thus, f((Id+L)s−1( W )) > r = f(W ), which is a contradiction. So L(W ) = rW ,�W �2 W ∈ Pk ∩ VMkV and I(W ) = I(V ). In order to complete this proof, we must show that r is the spectral radius of L with geometric multiplicity 1. Define L1 : VMkV → VMkV as L1(x) = 1 r M +L(MxM)M+, where M ∈ Z is such that M2 = W . Notice that L1 is a positive map such that L1(V ) = V . Next, if A is an eigenvector of L associated to some eigenvalue α then M+AM+ is an eigenvector of L1 associated to α/r. By lemma 2.5, �αr � ≤ 1. Hence, �α� ≤ r and the spectral radius of L is r. Finally, assume W2 ∈ VMkV is a Hermitian eigenvector of L associated to r. If W2 and W are linear independent then there is 0 ≠ µ ∈ R such that W − µW2 ∈ Pk and rank(W − µW2) < rank(W ), by lemma 2.2. Thus, L(V ′ MkV ′) ⊂ V ′ MkV ′ , where V ′ is the orthogonal projection onto I(W − µW2), by corollary 2.4. This contradicts the irreducibility of L, since I(W − µW2) ≠ I(W ) and V ′ ≠ V . Therefore, W2 and W are linear dependent. Since L preserves Hermiticity and r > 0 then every eingevector of L associated to r is a linear combination of Hermitian eigenvectors of L associated to r, thus the geometric multiplicity of r is 1. Theorem 2.8. Let L : VMkV → VMkV be a positive map. The spectral radius r of L is an eigenvalue of L associated to some Z ∈ Pk ∩ VMkV . 20 Proof. Consider the sequence of irreducible positive maps Ln : VMkV → VMkV defined by 1Ln(x) = L(x) + n tr(xV )V converging to L : VMkV → VMkV . Let Zn ∈ Pk ∩ VMkV be the unique eigenvector of Ln associated to the spectral radius rn of Ln satisfying �Zn�2 = 1, where � ⋅ �2 is the spectral norm, by theorem 2.7. Notice that {X ∈ Pk ∩VMkV, �X�2 = 1} is a compact set, therefore there is a subsequence (Znk )k∈N converging to some Z ∈ {X ∈ Pk ∩ VMkV, �X�2 = 1}. Finally, since the spectral radius changes continuosly with a matrix then lim = r, k→∞ rnk where r is the spectral radius of L. Thus, L(Z) = lim Lnk (Znk ) = lim Znk = rZ. k→∞ k→∞ rnk 2.2 Two related properties: completely reducibility and the decomposition property The main result of this section is the equivalence of the next two properties for a self-adjoint positive map L : VMkV → VMkV (proposition 2.13). Definition 2.9. (Completely Reducible Maps): A positive map L : VMkV → VMkV is called completely reducible, if there are orthogonal projections V1, . . . , Vs ∈ Mk such that ViVj = 0 (i ≠ j), ViV = Vi, VMkV = V1MkV1 ⊕ . . . ⊕ VsMkVs ⊕R, R ⊥ V1MkV1 ⊕ . . . ⊕ VsMkVs and (1) L(ViMkVi) ⊂ ViMkVi, (2) L� is irreducible, ViMkVi (3) L�R ≡ 0. Definition 2.10. Let L : VMkV → VMkV be a self-adjoint positive map. We say that L has the decomposition property if for every γ ∈ Pk ∩VMkV such that L(γ) = λγ, λ > 0 and V1 ∈ Mk is the orthogonal projection onto I(γ) then L�R ≡ 0, where R = (V −V1)MkV1 ⊕V1Mk(V −V1). Notice that R is the orthogonal complement of V1MkV1 ⊕ (V − V1)Mk(V − V1) in VMkV . Lemma 2.11. Let L : VMkV → VMkV be a self-adjoint positive map. L is irreducible if and only if the largest eigenvalue has multiplicity 1 with respect to an eigenvector γ ∈ Pk ∩VMkV such that I(γ) = I(V ). 21 Proof. Since L is self-adjoint, the eigenvalues of L are real numbers. Since L : VMkV → VMkV is a positive map, by theorem 2.8, the spectral radius λ is an eigenvalue and there exists γ ∈ Pk ∩ VMkV such that L(γ) = λγ. Therefore the spectral radius is the largest eigenvalue of L. Since L is irreducible, the multiplicity of λ is 1, by theorem 2.7. Let V1 ∈ Mk be the orthogonal projection onto I(γ). Notice that I(V1) ⊂ I(V ). By the previous corollary L(V1MkV1) ⊂ V1MkV1. Since L is irreducible then V1 = V and I(γ) = I(V ). For the converse, if L(V1MkV1) ⊂ V1MkV1, I(V1) ⊂ I(V ) then the positive map L : V1MkV1 → V1MkV1 has an eigenvector γ ′ ∈ Pk ∩ V1MkV1, by theorem 2.8. If I(V1) ≠ I(V ) then I(γ ′) ≠ I(γ) and γ ′ is not a multiple of γ. Since the multiplicity of the largest eigenvalue is 1 then γ ′ is associated to a different eigenvalue. Thus, γ ′ is orthogonal to γ, since L is self-adjoint. However, γ ′ and γ are positive semidefinite Hermitian matrices and I(γ ′) ⊂ I(V1) ⊂ I(V ) = I(γ), thus they can not be orthogonal. Thus, I(V1) = I(V ) and V1 = V , and L is irreducible. Lemma 2.12. Let L : VMkV → VMkV be a self-adjoint positive map. Let us assume that L has the decomposition property (definition 2.10). Let V ′ MkV ′ ⊂ VMkV be such that L(V ′ MkV ′) ⊂ V ′ MkV ′ then L�V ′ MkV ′ also has the decomposition property. Proof. Let γ ∈ Pk ∩V ′ MkV ′ be such that L(γ) = λγ, λ > 0. Since L : VMkV → VMkV has the decomposition property (definition 2.10) then L�R ≡ 0, where R = (V −V1)MkV1⊕V1Mk(V −V1) and V1 ∈ Mk is the orthogonal projection such that I(V1) = I(γ). Notice that I(V1) = I(γ) ⊂ I(V ′) ⊂ I(V ). Consider now R ′ = (V ′ − V1)MkV1 ⊕ V1Mk(V ′ − V1). Since (V ′ − V1)MkV1 = (V − V1)(V ′ − V1)MkV1 ⊂ (V − V1)MkV1 and V1Mk(V ′ − V1) = V1Mk(V ′ − V1)(V − V1) ⊂ V1Mk(V − V1) then R ′ ⊂ R and L�R ′ ≡ 0. Thus, L : V ′ MkV ′ → V ′ MkV ′ has the decomposition property. Proposition 2.13. If L : VMkV → VMkV is a self-adjoint positive map then L has the decomposition property if and only if L is completely reducible. Moreover, the orthogonal projections V1, . . . , Vs in definition 2.9 are unique and s ≥ the multiplicity of the largest eigen­ value of L. Proof. First, suppose that L has the decomposition property and let us prove that L is completely reducible by induction on the rank of V . Notice that if rank(V ) = 1 then dim(VMkV ) = 1 and L is irreducible on VMkV . Thus, L is completely reducible by def­ inition 2.9. Let us assume that rank(V ) > 1. Since L is a positive map then S = {γ� 0 ≠ γ ∈ Pk ∩VMkV, L(γ) = λγ, λ > 0} ≠ ∅, theorem 2.8. Let γ ∈ S be such that rank(γ) = min{rank(γ ′)� γ ′ ∈ S}. By corollary 2.4, L(V1MkV1) ⊂ V1MkV1, where V1 is the orthogonal projection onto I(γ). Now, if L� is not irreducible then there exists V ′ MkV ′ ⊂ V1MkV1 with rank(V ′) 0. However, rank(δ) ≤ rank(V1 ′) < rank(V1) = rank(γ). This contradicts the choice of γ. Thus, L� isV1MkV1 irreducible. Now, if rank(V1) = rank(V ) then V1 = V and L�VMkV is irreducible. Therefore, L : VMkV → VMkV is completely reducible by definition 2.9. Next, suppose rank(V1) < rank(V ). Since L(V1MkV1) ⊂ V1MkV1 and L is self-adjoint then L((V1MkV1)⊥) ⊂ (V1MkV1)⊥. Therefore, tr(L(V − V1)V1) = 0. Since L(V − V1) and V1 are positive semidefinite then I(L(V −V1)) ⊂ I(V −V1). By lemma 2.3, L((V −V1)Mk(V −V1)) ⊂ (V − V1)Mk(V − V1). Notice that L�(V −V1)Mk(V −V1) is a self-adjoint positive map with the decomposition property by lemma 2.12. Since rank(V − V1) < rank(V ), by induction on the rank, L�(V −V1)Mk(V −V1) is completely reducible. Thus, there are orthogonal projections V2, . . . , Vs ∈ Mk satisfying ViVj = 0 (i ≠ j), Vi(V − V1) = Vi (i ≥ 2), (V − V1)Mk(V − V1) = V2MkV2 ⊕ . . . ⊕ VsMkVs ⊕ R̃ with R̃ ⊥ V2MkV2 ⊕ . . . ⊕ VsMkVs, L� is irreducible for 2 ≤ i ≤ s and L�̃ ≡ 0.ViMkVi R Since L has the decomposition property then VMkV = V1MkV1 ⊕(V −V1)Mk(V −V1)⊕R, where L�R ≡ 0 and R ⊥ V1MkV1 ⊕ (V − V1)Mk(V − V1). Thus, we obtained VMkV = V1MkV1 ⊕ V2MkV2 ⊕ . . . ⊕ VsMkVs ⊕ R̃⊕R such that L�ViMkVi is irreducible for 1 ≤ i ≤ s and L�R̃⊕R ≡ 0. Notice that ViVj = 0, for 2 ≤ i ≠ j ≤ s and V1Vi = 0, for 2 ≤ i ≤ s, because I(Vi) ⊂ I(V − V1). Notice that R̃ ⊥ V2MkV2 ⊕ . . .⊕VsMkVs and R̃ ⊥ V1MkV1, because R̃ ⊂ (V −V1)Mk(V −V1). Therefore R̃ ⊕R ⊥ V1MkV1 ⊕ V2MkV2 ⊕ . . . ⊕ VsMkVs and L�R̃⊕R ≡ 0. Thus, L is completely reducible. For the converse, let us assume that L is completely reducible and let us prove that L has the decomposition property. Thus, VMkV = V1MkV1 ⊕ . . . ⊕ VsMkVs ⊕R, R ⊥ V1MkV1 ⊕ . . . ⊕ VsMkVs, L(ViMkVi) ⊂ ViMkVi, L�ViMkVi is irreducible and L�R ≡ 0. Assume L(γ ′) = λγ ′ , λ > 0 and γ ′ ∈ Pk ∩ VMkV and let V ′ ∈ Mk be the orthogonal projection onto I(γ ′). By corollary 2.4, we have L(V ′ MkV ′) ⊂ V ′ MkV ′ . Notice that, γ ′ = γ1 ′ + . . . + γs ′ , where γi ′ ∈ ViMkVi. Now, since I(γi′) ⊂ I(Vi) and I(Vi) ⊥ I(Vj ), for i ≠ j, then each γi ′ ∈ Pk. Since each ViMkVi is an invariant subspace of L then we also conclude that L(γ ′) = λγ ′ . Note that, not for every i, one has γ ′ = 0. Assume, without loss of generality, that i γ ′ = γ ′ i + . . . + γ ′ and γ ′ ≠ 0, for 1 ≤ i ≤ m ≤ s. i 1 m i Now, if for some 1 ≤ i ≤ m, I(γi′) ≠ I(Vi) then L�ViMkVi is not irreducible, by corollary 2.4, which is a contradiction. Therefore, I(γi′) = I(Vi) for 1 ≤ i ≤ m and V1 + ⋅ ⋅ ⋅ + Vm = V ′ . 23 Next, VMkV = V ′ MkV ′⊕(V −V ′)Mk(V −V ′)⊕R ′ , where R ′ = (V −V ′)MkV ′⊕V ′ Mk(V − V ′). Notice that R ′ ⊥ V ′ MkV ′ ⊕ (V − V ′)Mk(V − V ′). Now, V1MkV1 ⊕. . .⊕VmMkVm ⊂ V ′ MkV ′ and Vm+1MkVm+1 ⊕. . .⊕VsMkVs ⊂ (V −V ′)Mk(V − V ′), therefore R ′ ⊥ V1MkV1 ⊕ . . . ⊕ VsMkVs and R ′ ⊂ R. Therefore, L�R ′ ≡ 0 and L has the decomposition property by definition 2.10. Finally, if L : VMkV → VMkV is a self-adjoint completely reducible map then the non­ null eigenvalues of L are the non-null eigenvalues of L� . Since L� is irreducible ViMkVi ViMkVi then the multiplicity of the largest eigenvalue is 1 by lemma 2.11. Therefore each L�ViMkVi has at most one largest eigenvalue of L. Thus, s ≥ the multiplicity of the largest eigenvalue of L : VMkV → VMkV . Now, if L(V ′′ MkV ′′) ⊂ V ′′ MkV ′′ and L�V ′′ MkV ′′ is irreducible then by lemma 2.11, there is γ ′′ ∈ Pk ∩ V ′′ MkV ′′ such that L(γ ′′) = λγ ′′ , λ > 0 and I(γ ′′) = I(V ′′). As we noticed in the second part of this proof, there is ViMkVi ⊂ V ′′ MkV ′′ (V ′′ is a sum of some Vi’s). Since L(ViMkVi) ⊂ ViMkVi then L�V ′′ MkV ′′ is irreducible if and only if V ′′ = Vi, for some 1 ≤ i ≤ s. 24 Chapter 3 Completely Reducible Maps in Quantum Information Theory The results of this chapter were published in [14]. Let us identify the tensor product space Mk ⊗Mm with Mkm, via Kronecker product (i.e., if C = (cij ) ∈ Mk and B ∈ Mm then C ⊗B = (cij B) ∈ Mkm). Let A ∈ Mk ⊗Mm ≃ Mkm be a Hermitian matrix. We can write A = ∑n i=1 Ai ⊗Bi, where Ai,Bi are Hermitian matrices for every i. Let FA : Mm → Mk be FA(X) = ∑i n =1 tr(BiX)Ai and GA : Mk → Mm be GA(X) = ∑i n =1 tr(AiX)Bi. These maps are adjoint with respect to the trace inner product (Since Ai,Bi are Hermitian matrices then FA(Y ∗) = FA(Y )∗, for every Y ∈ Mm. Notice that if X ∈ Mk and Y ∈ Mm then tr(A(X ⊗ Y ∗)) = tr(GA(X)Y ∗) = tr(XFA(Y ∗)) = tr(XFA(Y )∗)). Notice that if {γ1, . . . , γk2 } is an orthonormal basis of Mk ∑k2formed by Hermitian matrices then A = i=1 γi ⊗GA(γi). Moreover, if A is positive semidefinite then FA : Mm → Mk and GA : Mk → Mm are also positive maps, since 0 ≤ tr(A(X ⊗ Y )) = tr(GA(X)Y ) = tr(XFA(Y )), when X ∈ Pk and Y ∈ Pm. Thus, FA ○GA :Mk →Mk is a self-adjoint positive map. In Section 3.1, we prove that if A ∈ Mk ⊗Mm is positive under partial transposition or symmetric with positive coefficients or invariant under realignment then FA ○GA :Mk →Mk is completely reducible (theorems 3.2, 3.12 and 3.13). These are our main results and we shall apply them to Quantum Information Theory. In Section 3.2, we apply our main results to two problems in Quantum Information Theory. We reduce the separability problem to the weakly irreducible case. We provide a complete description of weakly irreducible PPT matrices. We also show that if FA ○GA : Mk →Mk is 25 completely reducible with eigenvalues equal to 1 or 0 then A is separable. We use this result in order to obtain a new proof of the following one: If Ck contains k mutually unbiased bases then Ck contains k + 1. This is our last application to Quantum Information Theory. We complete this chapter with some remarks on our main theorems and on the ap­ plications to Quantum Information Theory. We present a couple of examples of positive semidefinite Hermitian matrices A in Mk ⊗Mk ≃ Mk2 such that FA ○GA : Mk → Mk is not completely reducible. We also show that if A ∈ M2 ⊗M2 is symmetric with positive coefficients or invariant under realignment then A is positive under partial transposition, reinforcing the connection between these three types. Finally, we show that low tensor rank implies sepa­ rability and we connect the aforementioned applications to Quantum Information Theory to this last result (see subsection 3.3.2). Throughout this chapter we shall adopt the following notation: Ck is the set of column vectors with k complex entries. We shall also identify the tensor product space Ck ⊗Cm with Ckm, via Kronecker product (i.e. if v = (vi) ∈ Ck,w ∈ Cm then v ⊗w = (viw) ∈ Ckm). The identification of the tensor product space Ck ⊗Cm with Ckm and the tensor product space Mk ⊗Mm with Mkm, via Kronecker product, allow us to write (v⊗w)(r⊗s)t = vrt ⊗wst, where v⊗w ∈ Ck ⊗Cm is a column, (v⊗w)t its transpose and v, r ∈ Ck and w, s ∈ Cm. Therefore if x, y ∈ Ck ⊗Cm ≃ Ckm we have xyt ∈ Mk ⊗Mm ≃ Mkm. Here, tr(A) denotes the trace of a matrix A, A stands for the matrix whose entries are aij, where aij is the complex conjugate of the entry aij of A and At stands for the transpose of A. We shall consider the usual inner product in Mk, ⟨A, B⟩ = tr(AB ∗), and the usual inner product in Ck, ⟨x, y⟩ = xty. If A = ∑n i=1 Ai ⊗Bi, we shall denote by At2 the matrix ∑n i=1 Ai ⊗Bi t, which is called the partial transposition of A. The image (or the range) of the matrix A ∈ Mk in Ck shall be denoted by I(A). 3.1 Main Theorems: The Complete Reducibility of FA ○GA :Mk →Mk 3.1.1 Main Theorem for PPT matrices Definition 3.1. (PPT matrices) Let A = ∑n i=1 Ai ⊗ Bi ∈ Mk ⊗Mm ≃ Mkm be a positive semidefinite Hermitian matrix. We say that A is positive under partial transposition or simply PPT, if At2 = Id ⊗ (⋅)t(A) = ∑n i=1 Ai ⊗Bi t is positive semidefinite. Theorem 3.2. Let A ∈ Mk ⊗Mm ≃ Mkm, A ∈ Pkm. If A is PPT then FA ○GA : Mk →Mk is completely reducible. 26 Proof. Let γ ∈ Pk ∩VMkV be such that FA(GA(γ)) = λγ, λ > 0. Let V1 ∈ Mk be the orthogonal projection onto I(γ). Let W1 ∈ Mm be the orthogonal projection onto I(GA(γ)). By lemma 2.3, we have GA(V1MkV1) ⊂ W1MmW1 and FA(W1MmW1) ⊂ V1MkV1. If V2 = Id − V1 and W2 = Id −W1 then A = ∑2 ).i,j,r,s=1(Vi ⊗Wj )A(Vr ⊗Ws Notice that tr(A(V1 ⊗W2)) = tr(GA(V1)W2) = 0. Thus, A(V1 ⊗W2) = (V1 ⊗W2)A = 0, since A ∈ Pkm and V1 ⊗W2 ∈ Pkm. Notice that tr(A(V2 ⊗W1)) = tr(V2FA(W1)) = 0. Thus, A(V2 ⊗W1) = (V2 ⊗W1)A = 0, since A ∈ Pkm and V2 ⊗W1 ∈ Pkm. Therefore, A = ∑2 i,j=1(Vi ⊗Wi)A(Vj ⊗Wj ). Next, 0 = (A(V1 ⊗W2))t2 = (Id ⊗W2 t)At2 (V1 ⊗ Id) and 0 = tr((Id ⊗W2 t)At2 (V1 ⊗ Id)) = tr(At2 (V1 ⊗ W2 t)). Since A is PPT then At2 is positive semidefinite and At2 (V1 ⊗ W2 t) = (V1 ⊗W2 t)At2 = 0. Analogously, we obtain At2 (V2 ⊗W1 t) = (V2 ⊗W1 t)At2 = 0. Thus, At2 = ∑2 (Vi ⊗W t)At2 (Vj ⊗W t) and At2 = ∑2 (Vi ⊗W t)At2 (Vi ⊗W t). Hence, i,j=1 j i i=1 i i A = ∑i 2 =1(Vi ⊗Wi)A(Vi ⊗Wi). Notice also that if X ∈ R = V1MkV2 ⊕ V2MkV1, which is the orthogonal complement of V1MkV1 ⊕ V2MkV2 in Mk, then GA(X) = 0 and FA ○GA�R ≡ 0. Thus, FA ○GA :Mk →Mk is a self-adjoint positive map with the decomposition property (definition 2.10). By proposition 2.13, FA ○GA :Mk →Mk is completely reducible. 3.1.2 Main Theorems for SPC matrices and Matrices Invariant under Realignment In order to obtain our main theorems for SPC matrices and matrices invariant under realign­ ment, we need some definitions and some preliminary results. Definition 3.3. Let {e1, . . . , ek} be the canonical basis of Ck. t t(1) Let T = ∑k ⊗ This matrix satisfies Ta ⊗ b = b ⊗ a,i,j=1 eiej ej ei ∈ Mk ⊗ Mk ≃ Mk2 . (a ⊗ b)tT = (b ⊗ a)t, for every a, b ∈ Ck, where a ⊗ b is a column vector in Ck2 and (a ⊗ b)t is its transpose. This matrix is usually called the flip operator (see [45]). (2) Let u = ∑i k =1 ei ⊗ ei ∈ Ck ⊗Ck. (3) Let F :Mk → Ck ⊗Ck, F (∑i n =1 aib t i) = ∑ n i=1 ai ⊗ bi. 27 Remark 3.4. Recall that F is an isometry, i.e., F (A)tF (B) = tr(AB ∗), for every A, B ∈ Mk, where F (A), F (B) ∈ Ck ⊗Ck and F (B) is the conjugation of the column vector F (B). We also have tr(F −1(v)F −1(w)∗) = vtw, for every v, w ∈ Ck2 (see [41]). Definition 3.5. Let S :Mk ⊗Mk →Mk ⊗Mk be defined by n n S(∑Ai ⊗Bi) =∑F (Ai)F (Bi)t , i=1 i=1 where F (Ai) ∈ Ck ⊗Ck is a column vector and F (Bi)t is a row vector (definition 3.3). This map is usually called the “realignment map" (see [18, 40,41]). Definition 3.6. (SPC matrices) Let A ∈ Mk ⊗Mk ≃ Mk2 be a positive semidefinite Hermi­ tian matrix. We say that A is symmetric with positive coefficients or simply SPC, if S(At2 ) is a positive semidefinite Hermitian matrix. Remark 3.7. The name symmetric with positive coefficients (SPC) is justified by proposition 3.33: If A ∈ Pk2 then A is SPC if and only if A has the following symmetric Hermitian Schmidt decomposition (definition 3.16) with positive coefficients: ∑n i=1 λiγi ⊗γi, with λi > 0, for every i. Definition 3.8. (Matrices Invariant under Realignment) Let A ∈ Mk ⊗Mk be a positive semidefinite Hermitian matrix. We say that A is invariant under realignment if A = S(A). Examples 3.9. a) Since Id ⊗ Id is invariant under partial transposition, (Id ⊗ Id)t2 = Id ⊗ Id, then Id ⊗ Id is PPT. Since S(Id ⊗ Id) = uut, by definitions 3.3 and 3.5, then Id ⊗ Id is also SPC. b) Since uut = ∑k t ⊗ eiet then S(uut) = ∑k t ⊗ ej et = Id ⊗ Id. Observe that i,j=1 eiej j i,j=1 eiei j tS(Id ⊗ Id + uut) = uut + Id ⊗ Id and Id ⊗ Id + uu is positive semidefinite. Thus, Id ⊗ Id + uut is invariant under realignment. t tc) Since T = ∑i,j k =1 eiej ⊗ ej ei then S(T ) = T . Since the eigenvalues of T are 1 and −1 then Id ⊗ Id − T is positive semidefinite. Hence, Id ⊗ Id + uut − T is invariant under realignment. Lemma 3.10. (Properties of the Realignment map) Let S : Mk ⊗Mk → Mk ⊗Mk be the realignment map defined in 3.5. Let v, vi,wi ∈ Ck ⊗Ck, V, W, M, N ∈ Mk. Then (1) S(At2 )v = F ○ FA ○ F ∗(v) t(2) S(∑n ) = ∑i n =1 F −1(vi)⊗ F −1(wi)i=1 viwi (3) S2 = Id :Mk ⊗Mk →Mk ⊗Mk (4) S((V ⊗W )A(M ⊗N)) = (V ⊗M t)S(A)(W t ⊗N) (5) S(AT )T = At2 28 (6) S(At2 ) = S(A)T (7) S(AT ) = S(A)t2 Proof. Let A = ∑n i=1 Ai ⊗Bi. Notice that At2 = ∑n i=1 Ai ⊗Bi t and S(At2 )v = ∑n i=1 F (Ai)F (Bi t)tv. By remark 3.4, since v = F (F −1(v)) then F (Bt)tv = tr(BtF −1(v)t) = tr(BiF −1(v)). There­ fore, S(At2 )v = F (∑n i=1 Aitr(BiF −1(v))) = F ○ i FA ○ F −1(v i ). Since F is an isometry then F −1 = F ∗, also by the same remark, and item 1 is proved. Next, since S(abt ⊗ cdt) = act ⊗ bdt for every a, b, c, d ∈ Ck then S2(abt ⊗ cdt) = abt ⊗ cdt. Since {abt ⊗ cdt , a, b, c, d ∈ Ck} is a set of generators of Mk ⊗Mk then item 3 is proved. By definiton 3.5, item 2 is also proved. In order to prove the other properties, since both sides of the equations are linear on A, we just need to prove for A = abt ⊗ cdt, where a, b, c, d ∈ Ck. Now, S((V ⊗W )(abt ⊗ cdt)(M ⊗N)) = S((V a ⊗Wc)(M tb ⊗N td)t). By item (2), this is equal to F −1(V a⊗Wc)⊗F −1(M tb⊗N td) = (V actW t)⊗(M tbdtN) = (V ⊗M t)(act ⊗bdt)(W t ⊗ N) = (V ⊗M t)S(A)(W t ⊗N). Thus, item 4 is proved. The other properties are also straightforward. Just recall that S(abt ⊗ cdt) = act ⊗ bdt, (abt ⊗ cdt)T = adt ⊗ cbt, T (abt ⊗ cdt) = (cbt ⊗ adt) and (abt ⊗ cdt)t2 = abt ⊗ dct, for every a, b, c, d ∈ Ck. Lemma 3.11. Let A ∈ Mk ⊗Mk be a Hermitian matrix. √ a) If S(At2 ) ∈ Pk2 and FA ○GA(γ) = λγ then GA(γ) = λγ. √ b) If S(A) ∈ Pk2 and FA ○GA(γ) = λγ then GA(γ) = λγt. Proof. Since S(At2 ) ∈ Pk2 then FA :Mk →Mk is a self-adjoint linear transformation with non negative eigenvalues, by item 1 of lemma 3.10 and by remark 3.4. Since A is Hermitian, GA is the adjoint of FA and FA = GA. Therefore, FA (γ) = λγ if and only if G2 (γ) = λγ if√ ○GA A and only if GA(γ) = λγ, because GA has only non negative eigenvalues. Thus, item a) is proved. Now, since At2 is Hermitian and (At2 )t2 = A then, by item a), FAt2 ○GAt2 (γ) = λγ if and √ only if GAt2 (γ) = λγ. But FAt2 ○GAt2 (X) = FA ○GA(X) and GAt2 (X) = GA(X)t. Theorem 3.12. Let A ∈ Mk ⊗Mk ≃ Mk2 , A ∈ Pk2 . If A is SPC then FA ○GA : Mk →Mk is completely reducible. 29 Proof. By definition 3.6, S(At2 ) ∈ Pk2 . Let γ ∈ Pk ∩ VMkV be such that FA(GA(γ)) = λ2γ, λ > 0. By item a) of lemma 3.11, GA(γ) = λγ. Thus, FA(γ) = λγ. Let V1 ∈ Mk be the orthogonal projection onto I(γ). By lemma 2.3, we have GA(V1MkV1) ⊂ V1MkV1 and FA(V1MkV1) ⊂ V1MkV1. If V2 = Id − V1 then A = ∑2 (Vi ⊗ Vj )A(Vr ⊗ Vs).i,j,r,s=1 Notice that tr(A(V1 ⊗ V2)) = tr(GA(V1)V2) = 0. Thus, A(V1 ⊗ V2) = (V1 ⊗ V2)A = 0, since A ∈ Pk2 and V1 ⊗ V2 ∈ Pk2 . Notice that tr(A(V2 ⊗ V1)) = tr(V2FA(V1)) = 0. Thus, A(V2 ⊗ V1) = (V2 ⊗ V1)A = 0, since A ∈ Pk2 and V2 ⊗ V1 ∈ Pk2 . Therefore, A = ∑2 (Vi ⊗ Vi)A(Vj ⊗ Vj ).i,j=1 Next, 0 = (A(V1 ⊗ V2))t2 = (Id ⊗ V2 t)At2 (V1 ⊗ Id) and 0 = S((Id ⊗ V2 t)At2 (V1 ⊗ Id)) = (Id ⊗ V1 t)S(At2 )(V2 ⊗ Id), by item 4 of lemma 3.10. Now, 0 = tr((Id ⊗ V1 t)S(At2 )(V2 ⊗ Id)) = tr(S(At2 )(V2 ⊗ V1 t)). Since S(At2 ) ∈ Pk2 then S(At2 )(V2 ⊗ V1 t) = (V2 ⊗ V1 t)S(At2 ) = 0. Analogously, we obtain S(At2 )(V1 ⊗ V2 t) = (V1 ⊗ V2 t)S(At2 ) = 0. Thus, At2 = ∑i,j 2 =1(Vi ⊗ Vj t)At2 (Vj ⊗ Vi t) and S(At2 ) = ∑i,j 2 =1(Vi ⊗ Vj t)S(At2 )(Vj ⊗ Vi t) = ∑2 i=1(Vi ⊗ Vi t)S(At2 )(Vi ⊗ Vi t), by item 4 of lemma 3.10. So, At2 = S2(At2 ) = ∑2 (Vi ⊗ V t)S2(At2 )(Vi ⊗ V t) = ∑2 (Vi ⊗ V t)At2 (Vi ⊗ V t), by items i=1 i i i=1 i i 3 and 4 of lemma 3.10. Therefore, A = ∑2 i=1(Vi ⊗ Vi)A(Vi ⊗ Vi). Finally, notice that if X ∈ R = V1MkV2 ⊕ V2MkV1, which is the orthogonal complement of V1MkV1 ⊕ V2MkV2 in Mk, then GA(X) = 0 and FA ○GA�R ≡ 0. Thus, FA ○GA :Mk →Mk is a self-adjoint positive map with the decomposition property (definition 2.10). By proposition 2.13, FA ○GA :Mk →Mk is completely reducible. Theorem 3.13. Let A ∈ Mk ⊗Mk ≃ Mk2 , A ∈ Pk2 . If A is invariant under realignment then FA ○GA :Mk →Mk is completely reducible. Proof. Let γ ∈ Pk ∩ VMkV be such that FA(GA(γ)) = λ2γ, λ > 0. By item b) of lemma 3.11, GA(γ) = λγt. Thus, FA(γt) = λγ. Let V1 ∈ Mk be the orthogonal projection onto I(γ). By lemma 2.3, we have GA(V1MkV1) ⊂ V tMkV t and FA(V tMkV t) ⊂ V1MkV1. Now, if V2 = Id−V11 1 1 1 then A = ∑2 i,j,r,s=1(Vi ⊗ Vj t)A(Vr ⊗ Vs t). Notice that tr(A(V1 ⊗ V t)) = tr(GA(V1)V t) = 0. Thus, A(V1 ⊗ V t) = (V1 ⊗ V t)A = 0,2 2 2 2 since A ∈ Pk2 and V1 ⊗ V t ∈ Pk2 . Notice that tr(A(V2 ⊗ V t)) = tr(V2FA(V t)) = 0. Thus, 2 1 1 A(V2 ⊗ V t) = (V2 ⊗ V t)A = 0, since A ∈ Pk2 and V2 ⊗ V t ∈ Pk2 .1 1 1 Therefore, A = ∑2 (Vi ⊗ V t)A(Vj ⊗ V t).i,j=1 i j Next, S(A) = ∑i,j 2 =1 S((Vi ⊗ Vi t)A(Vj ⊗ Vj t)) = ∑i,j 2 =1(Vi ⊗ Vj t)S(A)(Vi ⊗ Vj t), by item 4 of lemma 3.10. Since A = S(A), we have A = ∑2 i,j=1(Vi ⊗V t)A(Vi ⊗V t) = ∑2 i=1(Vi ⊗V t)A(Vi ⊗V t).j j i i Finally, notice that if X ∈ R = V1MkV2 ⊕ V2MkV1, which is the orthogonal complement of V1MkV1 ⊕ V2MkV2 in Mk, then GA(X) = 0 and FA ○GA�R ≡ 0. Thus, FA ○GA :Mk →Mk is a self-adjoint positive map with the decomposition property (definition 2.10). By proposition 2.13, FA ○GA :Mk →Mk is completely reducible. 30 3.2 Applications to Quantum Information Theory Throughout the following subsection we shall assume that FA ○GA :Mk →Mk is completely reducible. This is a strong restriction. However, we know that if A is PPT or SPC or invariant under realignment then FA ○ GA : Mk → Mk is indeed completely reducible (theorems 3.2, 3.12, 3.13). Recall that a necessary condition for the separability of a matrix is to be PPT. 3.2.1 The Separability Problem We assume that FA ○ GA : Mk → Mk is completely reducible and we give applications to Quantum Information Theory. The first application is the reduction of the Separability Problem to the weakly irreducible case (corollary 3.20) and the second is proposition 3.21 which grants the separability of A, if FA ○GA :Mk →Mk has only eigenvalues 1 or 0. Throughout the next section we present our last application concerning mutually unbiased bases using this proposition 3.21 for a matrix invariant under realignment (see proposition 3.22 and theorem 3.28). We begin this section with a simple lemma that provides an equivalent way to prove that FA ○GA :Mk →Mk is completely reducible. Lemma 3.14. Let A ∈ Mk ⊗Mm ≃ Mkm, A ∈ Pkm. Thus, FA ○GA : Mk → Mk is completely reducible if and only if for every γ ∈ Pk such that FA ○ GA(γ) = λγ, λ > 0, we have A = (V1 ⊗W1)A(V1 ⊗W1)+ (Id −V1 ⊗ Id −W1)A(Id −V1 ⊗ Id −W1), where V1 ∈ Mk,W1 ∈ Mm are orthogonal projections onto I(γ), I(GA(γ)), respectively. Proof. Suppose FA ○GA :Mk →Mk is completely reducible then FA ○GA :Mk →Mk has the decomposition property (definition 2.10) by proposition 2.13. If γ ∈ Pk is such that FA ○GA(γ) = λγ, λ > 0, then Mk = V1MkV1 ⊕(Id−V1)Mk(Id−V1)⊕R, where R ⊥ V1MkV1 ⊕(Id −V1)Mk(Id −V1) and FA ○GA�R ≡ 0, where V1 ∈ Mk is the orthogonal projection onto I(γ). Next, let W1 ∈ Mm be the orthogonal projection onto the I(GA(γ)). By lemma 2.3, we have GA(V1MkV1) ⊂ W1MmW1, because GA is a positive map, since A ∈ Pkm. Now, ⟨GA(Id − V1),GA(γ)⟩ = ⟨Id − V1, FA ○ GA(γ)⟩ = λ⟨Id − V1, γ⟩ = 0. Since GA(Id − V1) and GA(γ) are positive semidefinite then I(GA(Id − V1)) ⊥ I(GA(γ)) = I(W1). Thus, I(GA(Id − V1)) ⊂ I(Id −W1). Again by lemma 2.3, we have GA((Id − V1)Mk(Id − V1)) ⊂ (Id −W1)Mm(Id −W1). 31 Next, since FA ○GA�R ≡ 0 and FA,GA are adjoint maps then GA�R ≡ 0. Let {γ1, . . . , γr} be an orthonormal basis of V1MkV1 formed by Hermitian matrices, {δ1, . . . , δs} be an orthonormal basis of (Id−V1)Mk(Id−V1) formed by Hermitian matrices and {α1, . . . , αt} be an orthonormal basis of R formed by Hermitian matrices. Then A = ∑i r =1 γi ⊗GA(γi) + ∑i s =1 δi⊗GA(δi)+∑i s =1 αi⊗GA(αi). Since GA(αi) = 0 then A = ∑i r =1 γi⊗GA(γi)+∑i s =1 δi⊗GA(δi). Since {γ1, . . . , γr} ⊂ V1MkV1 then GA(γi) ∈ W1MmW1 and since {δ1, . . . , δs} ⊂ (Id − V1)Mk(Id − V1) then GA(δi) ∈ (Id −W1)Mm(Id −W1). Therefore, (V1 ⊗W1)A(V1 ⊗W1) = ∑i r =1 γi ⊗ GA(γi), (Id − V1 ⊗ Id − W1)A(Id − V1 ⊗ Id − W1) = ∑i s =1 δi ⊗ GA(δi) and A = (V1 ⊗W1)A(V1 ⊗W1) + (Id − V1 ⊗ Id −W1)A(Id − V1 ⊗ Id −W1). For the converse, assume that if γ ∈ Pk is such that FA ○ GA(γ) = λγ, λ > 0 then A = (V1 ⊗W1)A(V1 ⊗W1)+ (Id −V1 ⊗ Id −W1)A(Id −V1 ⊗ Id −W1), where V1,W1 are orthogonal projections onto I(γ), I(GA(γ)), respectively. Let Mk = V1MkV1 ⊕ (Id − V1)Mk(Id − V1)⊕R, R ⊥ V1MkV1 ⊕ (Id − V1)Mk(Id − V1). Notice that GA�R ≡ 0 and FA ○GA�R ≡ 0. Therefore, FA ○GA :Mk →Mk has the decompo­ sition property (definition 2.10) and by proposition 2.13, FA ○GA : Mk → Mk is completely reducible. Definition 3.15. Let A ∈ Mk ⊗Mm ≃ Mkm, A ∈ Pkm. We say that A is weakly irreducible if for every orthogonal projections V1, V2 ∈ Mk and W1,W2 ∈ Mm such that V2 = Id−V1, W2 = Id−W1 and A = (V1 ⊗W1)A(V1 ⊗W1)+ (V2 ⊗W2)A(V2 ⊗W2), we obtain (V1 ⊗W1)A(V1 ⊗W1) = 0 or (V2 ⊗W2)A(V2 ⊗W2) = 0. Definition 3.16. A decomposition of a matrix A ∈ Mk ⊗Mm, ∑n i=1 λiγi ⊗ δi, is a Schmidt decomposition if {γi� 1 ≤ i ≤ n} ⊂ Mk, {δi� 1 ≤ i ≤ n} ⊂ Mm are orthonormal sets with respect to the trace inner product, λi ∈ R and λi > 0. Also, if γi and δi are Hermitian matrices for every i, then ∑n i=1 λiγi ⊗ δi is a Hermitian Schmidt decomposition of A. Proposition 3.17. Let A ∈ Mk ⊗Mm ≃ Mkm, A ∈ Pkm. Let ∑n i=1 λiγi ⊗ δi be a Hermitian Schmidt decomposition of A such that λ1 ≥ λ2 ≥ . . . ≥ λn > 0. If FA ○ GA : Mk → Mk is completely reducible then the following conditions are equivalent: (1) A is weakly irreducible, (2) s = 1 in definition 2.9 with L = FA ○GA :Mk →Mk, (3) λ1 > λ2 and I(γi) ⊂ I(γ1), I(δi) ⊂ I(δ1), for 1 ≤ i ≤ n. 32 http:Definition3.15 Proof. Notice that FA ○GA(γi) = λi 2γi for 1 ≤ i ≤ n. Thus, the largest eigenvalue of FA ○GA is λ2 1. By definition 2.9, Mk = V1MkV1 ⊕ . . . ⊕ VsMkVs ⊕ R, FA ○ GA�ViMkVi is irreducible and FA ○ GA�R ≡ 0, where R ⊥ V1MkV1 ⊕ . . . ⊕ VsMkVs. Since the non-null eigenvalues of FA ○ GA : Mk → Mk are eigenvalues of FA ○ GA�ViMk Vi , for 1 ≤ i ≤ s, then λ2 1 is the largest eigenvalue of some FA ○GA� . Without loss of generality we may assume that there exists ViMkVi 0 ≠ γ ∈ Pk ∩ V1MkV1 such that FA ○GA(γ) = λ1 2γ and I(γ) = I(V1), by lemma 2.11. Thus, by lemma 3.14, A = (V1 ⊗W1)A(V1 ⊗W1) + (Id − V1 ⊗ Id −W1)A(Id − V1 ⊗ Id −W1), where V1,W1 are orthogonal projections onto I(γ), I(GA(γ)), respectively. Firstly, let us assume that A is weakly irreducible, then or (V1 ⊗W1)A(V1 ⊗W1) = 0 or (Id − V1 ⊗ Id −W1)A(Id − V1 ⊗ Id −W1) = 0. Notice that if (V1 ⊗W1)A(V1 ⊗W1) = 0 then A = (Id −V1 ⊗ Id −W1)A(Id −V1 ⊗ Id −W1) and GA(γ) = 0, since γ ∈ V1MkV1. Therefore, 0 = FA○GA(γ) = λ1 2γ, which is a contradiction. Therefore (Id−V1 ⊗Id−W1)A(Id−V1 ⊗Id−W1) = 0 and A = (V1 ⊗W1)A(V1 ⊗W1). In this case, GA�(V1Mk V1)⊥ ≡ 0 and FA ○GA�(V1MkV1)⊥ ≡ 0. Thus, s = 1 in definition 2.9. Secondly, suppose that s = 1 in definition 2.9 then Mk = V1MkV1 ⊕ R, FA ○ GA�V1MkV1 is irreducible and FA ○ GA�R ≡ 0, where R = (V1MkV1)⊥. Thus, γi ∈ V1MkV1 for 1 ≤ i ≤ n, since FA ○ GA(γi) = λ2 i γi and FA ○ GA(Mk) = FA ○ GA(V1MkV1) ⊂ V1MkV1. By lemma 2.3, GA(V1MkV1) ⊂ W1MmW1, since I(GA(γ)) = I(W1). Thus, λiδi = GA(γi) ∈ W1MmW1 and I(δi) ⊂ I(W1). Next, since FA ○GA : V1MkV1 → V1MkV1 is irreducible then the multiplicity of the largest eigenvalue is 1 by lemma 2.11, thus λ2 1 > λ2 2 and λ1 > λ2. Moreover, γ must be a multiple of γ1, because FA ○GA(γ1) = λ2 1γ1. Thus, GA(γ) is also a multiple of δ1. Therefore, I(γi) ⊂ I(V1) = I(γ) = I(γ1) and I(δi) ⊂ I(W1) = I(GA(γ)) = I(δ1). Finally, let us assume that λ1 > λ2 and I(γi) ⊂ I(γ1), I(δi) ⊂ I(δ1), for 1 ≤ i ≤ n. Let Vj ′ ∈ Mk and Wj ′ ∈ Mm, j = 1, 2, be orthogonal projections such that V2 ′ = Id−V1 ′ , W2 ′ = Id−W1 ′ and A = (V1 ′ ⊗W1 ′)A(V1 ′ ⊗W1 ′) + (V2 ′ ⊗W2 ′)A(V2 ′ ⊗W2 ′). Thus, GA�V ′ MkV ′+V ′ MkV ′ ≡ 0 and 1 2 2 1 ○GA�V ′ MkV ′+V ′ MkV ′ ≡ 0.FA 1 2 2 1 Next, notice that GA(V ′ MkV ′) ⊂ W ′ MmW ′ and FA(W ′ MmW ′) ⊂ V ′ MkV ′ , j = 1, 2, since j j j j j j j j V1 ′ V2 ′ = 0 and W1 ′ W2 ′ = 0. Thus, FA ○GA(Vj ′ MkVj ′) ⊂ Vj ′ MkVj ′ , for j = 1, 2. Hence, the non-null eigenvalues of FA ○ GA : Mk → Mk are the non-null eigenvalues of FA ○GA�V ′ Mk V ′ , j = 1, 2. Without loss of generality, let us assume that λ2 1 is an eigenvalue j j of FA ○ GA�V1 ′ Mk V1 ′ . Since the multiplicity of λ1 2 is 1 (λ1 > λ2) then γ1 ∈ V1 ′ MkV1 ′ . Since I(γj ) ⊂ I(γ1) ⊂ I(V1) ⊥ I(V2), j = 1, 2, then (V ′ ⊗W ′)A(V ′ ⊗ W ′) = 0. Therefore, A is2 2 2 2 weakly irreducible. Proposition 3.18. Let A ∈ Mk ⊗Mm ≃ Mkm, A ∈ Pkm. If FA ○GA : Mk →Mk is completely reducible then A = ∑s i=1(Vi ⊗Wi)A(Vi ⊗Wi) such that 33 (1) V1, . . . , Vs ∈ Mk are orthogonal projections such that ViVj = 0 (2) W1, . . . ,Ws ∈ Mm are orthogonal projections such that WiWj = 0 (3) (Vi ⊗Wi)A(Vi ⊗Wi) is weakly irreducible and non-null for every i. (4) s ≥ multiplicity of the largest eigenvalue of FA ○GA :Mk →Mk. Proof. Since FA ○GA :Mk →Mk is completely reducible then Mk = V1MkV1 ⊕. . .⊕VsMkVs ⊕R, FA ○GA(ViMkVi) ⊂ ViMkVi, FA ○GA�ViMkVi is irreducible, FA ○GA�R ≡ 0 and s ≥ multiplicity of the largest eigenvalue of FA ○GA :Mk →Mk, by proposition 2.13. By lemma 2.11, there is γ1 j ∈ Pk ∩ Vj MkVj , 1 ≤ j ≤ s, such that γ1 j is an eigenvector of FA ○GA : Vj MkVj → Vj MkVj associated to the unique largest eigenvalue and I(γ1 j ) = I(Vj ). Since GA is a positive map then GA(γ1 j ) ∈ Pm. By lemma 2.3, GA(Vj MkVj ) ⊂ Wj MkWj , where Wj is the orthogonal projection onto I(GA(γ1 j )). Notice that VjMkVj ⊥ ViMkVi, for i ≠ j, since ViVj = 0. Therefore, ⟨GA(γ1 j ),GA(γ1i )⟩ = ⟨γ1 j , FA ○GA(γ1i )⟩ = 0, for i ≠ j. Thus, WiWj = 0 for i ≠ j. Let {γ1 j , . . . , γr j j } be an orthonormal basis of Vj MkVj formed by Hermitian matrices. Let {δ1, . . . , δr} be an orthonormal basis of R formed by Hermitian matrices. Thus, ⋃s j=1{γ1 j , . . . , γr j j }∪ {δ1, . . . , δr} is an orthonormal basis of Mk formed by Hermitian matrices. Let Aj = γ1 j ⊗ GA(γj ) + . . . + γr j j ⊗GA(γrj j ). Thus, A = ∑j s =1 Aj + δ1 ⊗GA(δ1) + ⋅ ⋅ ⋅ + δr ⊗GA(δr).1 Now, since FA ○GA�R ≡ 0 and FA,GA are adjoint then GA�R ≡ 0 and A = ∑j s =1 Aj . Next, (Vj ⊗Wj )A(Vj ⊗Wj ) = Aj , since γl j ∈ Vj MkVj , GA(γl j) ∈ Wj MmWj , ViVj = 0 and WiWj = 0 for i ≠ j. Therefore, Aj ∈ Pkm. Notice that, FAj ○GAj �Vj MkVj = FA ○GA�Vj MkVj which is irreducible. Therefore Aj ≠ 0 for 1 ≤ j ≤ s. Next, Mk = Vj MkVj ⊕ (Vj MkVj )⊥ and FAj ○GAj ((Vj MkVj )⊥) = 0. Therefore, by definition 2.9, FAj ○GAj :Mk →Mk is completely reducible with s = 1 . Finally, by item 2 of proposition 3.17, Aj is weakly irreducible. Definition 3.19. (Separable Matrices) Let A ∈ Mk ⊗Mm. We say that A is separable if A = ∑n i=1 Ci ⊗Di such that Ci ∈ Mk and Di ∈ Mm are positive semidefinite Hermitian matrices for every i. Corollary 3.20. Let A be the matrix of proposition 3.18. Then A is separable if and only if each (Vi ⊗Wi)A(Vi ⊗Wi) is separable. Thus, for this type of A the Separability Problem is reduced to the weakly irreducible case. Proposition 3.21. Let A ∈ Mk ⊗Mm ≃ Mkm, A ∈ Pkm. If FA ○GA : Mk →Mk is completely reducible with all eigenvalues equal to 1 or 0 then there exists a unique Hermitian Schmidt decomposition of A, ∑n i=1 γi ⊗ δi, such that γi ∈ Pk, δi ∈ Pm. Therefore, A is separable. 34 Proof. Suppose the multiplicity of the eigenvalue 1 is n. Since FA ○ GA : Mk → Mk is completely reducible then there are orthogonal projections V1, . . . , Vs such that I(Vi) ⊥ I(Vj ), Mk = V1MkV1 ⊕ . . . ⊕ VsMkVs ⊕R, FA ○GA(ViMkVi) ⊂ ViMkVi, FA ○GA�ViMkVi is irreducible, FA ○GA�R ≡ 0 and s ≥ n, by definition 2.9 and proposition 2.13. Recall that each FA ○GA�ViMkVi has a unique largest eigenvalue, since FA ○GA� is irreducible by lemma 2.11. Moreover, ViMkVi the eigenvalues of FA ○GA� are 1 or 0. Thus, s = n and for each FA ○GA�ViMk Vi thereViMkVi exists a unique normalized eigenvector γi ∈ Pk such that FA ○GA(γi) = γi and I(γi) = I(Vi), by lemma 2.11. Note that I(γi) = I(Vi) ⊥ I(Vj ) = I(γj ), therefore γ1, . . . , γn are orthonormal. Complete this set to obtain an orthonormal basis {γ1, . . . , γn, γn+1, . . . , γk2 } of Mk formed by Hermitian matrices. Notice that FA ○ GA(γj ) = 0, for j > n. Since FA and GA are adjoint maps, GA(γj ) = 0 for j > n. Thus, A = γ1 ⊗GA(γ1)+ . . . + γk2 ⊗GA(γk2 ) = γ1 ⊗GA(γ1)+ . . . + γn ⊗GA(γn). Notice that ⟨GA(γi),GA(γj )⟩ = ⟨γi, FA ○GA(γj )⟩ = ⟨γi, γj ⟩, 1 ≤ i, j ≤ n, therefore GA(γ1), . . . ,GA(γn) are orthonormal too. Recall that GA is a positive map then GA(γi) ∈ Pm. Define δi = GA(γi). Finally, if ∑n i=1 γi ′ ⊗ δi ′ is a Hermitian Schmidt decomposition with γi ′ ∈ Pk, δ i ′ ∈ Pm then FA ○GA(γi′) = γi′ . Thus, FA ○GA(Vi ′ MkVi ′) ⊂ Vi ′ MkVi ′ , 1 ≤ i ≤ n, where Vi ′ is the orthogonal projection onto I(γ ′), by corollary 2.4. Notice that each FA �V ′ has one eigenvalue i ○GA i MkVi ′ equal to 1 and the others equal to 0. Thus, FA ○ GA�V ′ MkV ′ is irreducible by lemma 2.11. i i Now, each Vi ′ must be equal to some Vj , by proposition 2.13. Since each FA ○GA�Vj MkVj = FA ○GA�V ′ MkV ′ has only one eigenvalue equal to 1 then γi ′ is a i i multiple of γj , but both matrices are positive semidefinite Hermitian matrices and normalized then γi ′ = γj . Thus, each γi ′ is equal to some γj and this Hermitian Schmidt decomposition is unique. 3.2.2 Extension of Mutually Unbiased Bases In this subsection we obtain a new proof of the following theorem proved in [47]: If there is a set of k mutually unbiased bases of Ck then there exists another orthonormal basis which is mutually unbiased with these k bases. Our proof relies on proposition 3.22. We also proved that this additional basis is unique up to multiplication by complex numbers of norm 1. Proposition 3.22. Let A ∈ Mk ⊗Mk ≃ Mk2 , A ∈ Pk2 . If A is invariant under realignment and FA ○GA :Mk →Mk has n eigenvalues equal to 1 and the others 0 then a) there exists an orthonormal set {v1, . . . , vn} ⊂ Ck such that A = ∑i n =1 vivi t ⊗ vivi t . b) The orthonormal set of item a) is unique up to multiplication by complex numbers of norm one. 35 Proof. By theorem 3.13, FA ○GA : Mk → Mk is completely reducible. By proposition 3.21, there exists a unique Hermitian Schmidt decomposition of A , ∑n i=1 γi ⊗ δi, such that γi ∈ Pk, δi ∈ Pk for 1 ≤ i ≤ n. Notice that FA(GA(γi)) = γi, for every i. So, by item b) of lemma 3.11, GA(γi) = γt. Thus, γt = GA(γi) = δi. Therefore, A = ∑i n =1 γi ⊗γt is the unique Hermitian i i i Schmidt decomposition of A such that γi ∈ Pk for 1 ≤ i ≤ n. Let Vi be the orthogonal projection onto I(γi). Since {γ1, . . . , γn} is an orthonormal set and each γi ∈ Pk then I(Vi) ⊥ I(Vj ). Thus, (Vi ⊗ Vi t)A(Vi ⊗ Vi t) = γi ⊗ γi t . Now, let F (γi) = ri(see definition 3.3). By definition 3.5, ririt = S(γi ⊗ γi). Since γi is Hermitian then γi ⊗ γi = γi ⊗ γi t and ririt = S(γi ⊗ γi t) = S((Vi ⊗ Vi t)A(Vi ⊗ Vi t)). Next, by item 4 of lemma 3.10, S((Vi ⊗ V t)A(Vi ⊗ V t)) = (Vi ⊗ V t)S(A)(Vi ⊗ V t). Since i i i i S(A) = A then ririt = (Vi ⊗ V t)A(Vi ⊗ V t) = γi ⊗ γt i i i . t tTherefore, γi ⊗ γi t has rank 1 and γi has rank 1. Thus, γi = vivi and A = ∑i n =1 vivi t ⊗ vivi . Since tr(γiγj ) = δij then {v1, . . . , vn} is an orthonormal set. tFinally, suppose A = ∑n j=1 wj wj t ⊗ wj wj for another orthonormal set {w1, . . . ,wn}. Since ∑n i=1 γi ⊗ γi t is unique (such that γi ∈ Pk for 1 ≤ i ≤ n) then for each p there is q such that t =wpwp vqvq t. Therefore wp = cvq with �c� = 1. Definition 3.23. (Mutually Unbiased Bases) Let {v1, . . . , vk} and {w1, . . . ,wk} be or­ = 1thonormal bases of Ck. We say that they are mutually unbiased if �⟨vi,wj ⟩�2 k for every i, j. Definition 3.24. Let α = {v1, . . . , vk} be an orthonormal basis of Ck. Let us define Aα ∈ Mk ⊗Mk as Aα = ∑k i=1 vivi t ⊗ vivi t. Notice that Aα is invariant under realignment. Lemma 3.25. If α, β are orthonormal bases of Ck then they are mutually unbiased if and 1 tonly if AαAβ = AβAα = k uu (Recall the definition of u in 3.3). t tProof. Let α = {v1, . . . , vk}, β = {w1, . . . ,wk} and Aα = ∑i k =1 vivi t⊗vivi , Aβ = ∑j k =1 wj wj t⊗wj wj . Notice that AαAβ = ∑i,j k =1 viwj t ⊗ viwj t (vit wj )(vitwj ). t t = 1If α, β are mutually unbiased then for every i, j, we have (vi wj )(vi wj ) = �⟨vi,wj⟩�2 k . 1 t 1Therefore, AαAβ = ∑i,j k =1 viwj t ⊗ viw = uut, since u = ∑i k =1 vi ⊗ vi = ∑k j=1 wj ⊗wj .k j k tNow suppose that AαAβ = 1 uut. Therefore ∑k t ⊗ viw = 1 uut, where λij = k i,j=1 λij viwj j k �⟨vi,wj ⟩�2. Next, 1 Id ⊗ Id = S( 1 uut) = ∑i,j k =1 λij vivi t ⊗wj wj t . Notice that {vi ⊗wj � 1 ≤ i, j ≤ k}k k is an orthonormal basis of Ck ⊗ Ck. Therefore λij are the eigenvalues of k 1 Id ⊗ Id, thus 1λij = .k 36 Lemma 3.26. Let α1, . . . , αk+1 be orthonormal bases of Ck. If they are pairwise mutually unbiased then ∑k i= + 1 1 Aαi = Id ⊗ Id + uut ∈ Mk ⊗Mk. Proof. Since Aα1 , . . . ,Aαk+1 commute, by lemma 3.25, there is a common basis of Ck ⊗ Ck formed by orthonormal eigenvectors. Since Aα1 , . . . ,Aαk+1 are orthogonal projections and their pairwise multiplications are equal to √u √u t , by lemma 3.25, the intersection of their k k images is generated only by u. Notice that each Aαi has rank k. Thus, every Aαi can be written as √ u √u t +∑i(k−1) rlrl t, where r1, . . . , rk2−1, √u is a k k l=(i−1)(k−1)+1 k common orthonormal basis of eigenvectors. √u t +∑k2−1 u √u t u √u t +∑k2−1 tFinally, ∑k+1 Aαi = (k+1)√u rlrl t = k √ +√ riri = uut +Id⊗Id.i=1 l=1 i=1k k k k k k Remark 3.27. Adapting the proof of the previous lemma, we can show the following: If 1α1, . . . , αk+1 are pairwise mutually unbiased orthonormal bases of R2k then ∑k+1 Aαi = (Id ⊗i=1 2 Id + T + uut), where T is the flip operator (see definition 3.3). Theorem 3.28. If Ck contains k mutually unbiased bases then there exists another orthonor­ mal basis which is mutually unbiased with these k bases. This additional one is unique up to multiplication by complex numbers of norm one. Proof. Let α1, . . . , αk be orthonormal bases of Ck, which are pairwise mutually unbiased. Consider B = Id ⊗ Id + uut − (∑i k =1 Aαi ) ∈ Mk ⊗ Mk. Recall Aαi ∈ Mk ⊗ Mk ≃ Mk2 from definition 3.24. Since Aα1 , . . . ,Aαk are commuting orthogonal projections and their pairwise multiplica­ tions are equal to √u √u t , by lemma 3.25, then every Aαi , 1 ≤ i ≤ k, can be written as √u √u t +∑i(k−1) r k lrl k t, where r1, . . . , rk2−1, √u is a common orthonormal basis of eigenvec­ k k l=(i−1)(k−1)+1 k tors. Therefore, B = Id⊗Id+uut −(∑i k =1 Aαi ) = (k +1)√u √ut +∑l k = 2 1 −1 rlrl t −k √u √u t −∑l k = ( 1 k−1) rlrl t = k k k k √u √u t +∑k l= 2 k − ( 1 k−1)+1 rlrl t. Thus, B is an orthogonal projection with k eigenvalues equal to 1 k k and the others zero and BAαi = B = √u √ut , 1 ≤ i ≤ k.Aαi k k In order to complete the proof, we must show that B = Aαk+1 for some orthonormal basis αk+1 of Ck (definition 3.24). Since BAαi = Aαi B = √u √ut , 1 ≤ i ≤ k, then αk+1 is mutually unbiased with each αi, 1 ≤ i ≤ k, by lemma 3.25. k k Next, S(B) = S(Id ⊗ Id +uut −∑k i=1 Aαi ) = S(Id ⊗ Id +uut)−∑k i=1 S(Aαi ) = Id ⊗ Id +uut − ∑k i=1 Aαi = B, by item a) in example 3.9 and by definition 3.24. Thus B is invariant under realignment. 37 By item b) of proposition 3.33, we know that B has a Hermitian Schmidt decomposition ∑n i=1 λiγi ⊗γi t with λi > 0. Therefore, B = S(B) = ∑n i=1 λivivi t, where vi = F (γi). Since F is an tisometry, by remark 3.4, ∑n i=1 λivivi is a spectral decomposition of B and λi are the non-null eigenvalues of B. Then n = k and λi = 1, for 1 ≤ i ≤ k. Thus, ∑k i=1 γi ⊗ γi t is a Hermitian Schmidt decomposition of B and FB ○GB : Mk → Mk is FB ○ GB(X) = ∑i k =1 tr(γiX)γi, where γ1, . . . , γk are orthonormal eigenvectors of FB ○ GB associated to the eigenvalue 1. By proposition 3.22, there exists an orthonormal basis αk+1 of Ck such that B = Aαk+1 and this basis is unique up to multiplication by complex numbers of norm one. Remark 3.29. Assume α1, . . . , αk are pairwise mutually unbiased orthonormal bases of R2k 1and define B = 2 (Id⊗Id+T +uut)−(∑k i=1 Aαi ). We can repeat the proof of the previous theorem in order to obtain B = Aαk+1 , for some orthonormal basis αk+1 of C2k, since Id⊗Id+T +uut is invariant under realignment. The basis αk+1 is actually a basis of R2k (up to multiplication by complex numbers of norm 1), because B is also invariant under partial transposition. Thus, if R2k has k pairwise mutually unbiased bases then there exists another orthonormal basis which is mutually unbiased with these k bases. 3.3 Remarks 3.3.1 Some Remarks on our Main Theorems All the results within this subsetion were published in [13,14]. Below we present a couple of easy examples showing that FA ○ GA : Mk → Mk is not completely reducible in general (lemmas 3.30, 3.31). The assumption that A is PPT or SPC or invariant under realignment is essential in order to obtain the complete reducibility of FA ○GA :Mk →Mk (theorems 3.2, 3.12, 3.13). Thus, these three types of matrices are connected and we can ask the following question: Is it possible that every SPC matrix or every matrix invariant under realignment is PPT? The answer is YES in M2 ⊗M2 (see lemma 3.34) and NO in Mk ⊗Mk, k > 2 (see examples 3.36). Lemma 3.30. Let u ∈ Ck ⊗Ck, k ≥ 2, be the vector defined in 3.3 and A = uut ∈ Mk ⊗Mk. The linear transformation FA ○GA :Mk →Mk is not completely reducible. Proof. By definition 3.3, u = ∑k i=1 ei ⊗ei, where {e1, . . . , ek} is the canonical basis of Ck. Thus, t t t tA = uut = ∑k ⊗ eie and GA(X) = FA(X) = ∑k tr(eie X) = X t.i,j=1 eiej j i,j=1 eiej j Now, the identity map Id = FA ○GA : Mk → Mk has null kernel and every matrix is an eigenvector. Thus, Id :Mk →Mk does not have the decomposition property (definition 2.10) and FA ○GA is not completely reducible by proposition 2.13. 38 t i tLemma 3.31. Let k ≥ 3. Let v1, e3 ∈ Ck be such that v1 = (√1 , √ , 0, . . . , 0) and e3 = 2 2 (0, 0, 1, 0, . . . , 0). Consider v = v1 ⊗ v1 + e3 ⊗ e3 ∈ Ck ⊗Ck. Let A be the positive semidefinite Hermitian matrix A = vv t +S(vvt) ∈ Mk ⊗Mk. The map FA ○GA :Mk →Mk is not completely reducible. e3v1 t+v1e i(e3v1 t−v1e )3 3 t t =Proof. Let γ1 = v1v1 t, γ2 = √ t , γ3 = √ t , γ4 = e3e3. Notice that vv ∑i 4 =1 γi ⊗ γi t . 2 2 tNow, S(vvt) = V ⊗ V , where V = F −1(v) = v1v1 t + e3e3, by item 2 of lemma 3.10. Thus, A = ∑i 4 =1 γi ⊗ γi t + V ⊗ V . Since 0 = tr(γ1γ2) = tr(γ1γ3) = tr(γ1γ4) = tr(γ1V ) then GA(γ1) = γ1 t and FA(γ1t ) = γ1. Therefore FA ○ GA(γ1) = γ1. Next, 0 = tr(γ2γ1) = tr(γ2γ3) = tr(γ2γ4) = tr(γ2V ). Thus, GA(γ2) = γ2 t and FA(γ2t ) = γ2, thus FA ○GA(γ2) = γ2. Finally, notice that γ2 ∈ (Id − V1)MkV1 ⊕ V1Mk(Id − V1) = R, where V1 is the orthogonal projection onto I(γ1). Therefore FA ○GA�R ≠ 0. Thus, FA ○GA does not have the decom­ position property(definition 2.10) and FA ○ GA is not completely reducible by proposition 2.13. Lemma 3.32. Let A ∈ M2 ⊗M2 ≃ M4. Suppose A has a Hermitian Schmidt decomposition ∑m i=1 λiγi ⊗ γi i=1 λi det(γiwith λi > 0, for every i. If ∑m ) ≥ 0 then A is separable. Proof. If det(γi) = 0, for every i, then γi has rank 1 and γi ⊗ γi ∈ P4. Therefore each γi ⊗ γi is separable and ∑m i=1 λiγi ⊗ γi is separable too, since λi > 0 for every i . If there is i such that det(γi) ≠ 0 then we must have at least one i such that det(γi) > 0, because ∑m i=1 λi det(γi) ≥ 0 and λi > 0 for every i . Let us assume det(γ1) > 0. It means that the eigenvalues of γ1 are both positive or negative. Therefore γ1 ⊗γ1 is positive definite. Thus, we can assume that γ1 is positive definite. Let γ1 = N2 for some invertible N ∈ P2. Consider B = (N−1 ⊗N−1)A(N−1 ⊗N−1) = λ1Id ⊗ Id +∑m i=2 λiN−1γiN−1 ⊗N−1γiN−1. Now, since tr(γiγ1) = 0, for every i > 1, then γi has a positive and a negative eigenvalue. Thus, det(γi) < 0, for every i > 1. det(γi) ldet(γi)l det(γ1) ≥ 0 and det(γi ≥ ∑mNext, since λ1 +∑m i=2 λi ) < 0, for every i > 1, then λ1 i=2 λi det(γ1) ldet(γi)l ldet(γi)land B = (λ1 −∑m i=2 λi det(γ1) )Id ⊗ Id +∑m i=2 λi( det(γ1) Id ⊗ Id +N−1γiN−1 ⊗N−1γiN−1). Now, the smallest eigenvalue of N−1γiN−1 ⊗N−1γiN−1 is the product of the two distinct eigenvalues of N−1γiN−1 (since they have opposite signs), which is equal to det(N−1γiN−1) = det(γi) ldet(γi)l det(γ1) . Therefore, det(γ1) Id ⊗ Id +N−1γiN−1 ⊗N−1γiN−1 ∈ P4 and has tensor rank 2. Thus, ldet(γi)lby theorem 3.44, det(γ1) Id ⊗ Id +N−1γiN−1 ⊗N−1γiN−1 is separable, for every i, and B is separable. Therefore A is separable. 39 Proposition 3.33. Let A ∈ Mk ⊗Mk be a Hermitian matrix. a) S(At2 ) ∈ Pk2 if and only if there is a Hermitian Schmidt decomposition of A, ∑n i=1 λiγi ⊗ γi, such that λi > 0 for every i. b) S(A) ∈ Pk2 if and only if there is a Hermitian Schmidt decomposition of A, ∑i n =1 λiγi⊗γit , such that λi > 0 for every i. Proof. Since A is Hermitian then FA and GA are adjoint linear transformations. Therefore, FA ○ GA : Mk → Mk is a self-adjoint linear transformation with non negative eigenvalues. Moreover, the set of Hermitian matrices is left invariant by FA ○ GA : Mk → Mk. Thus, there exists an orthonormal basis of Hermitian matrices of Mk, {γ1, . . . , γk2 }, formed by eigenvectors of FA ○GA : Mk →Mk. Let {λ1 2, . . . , λ k 2 2 } be the corresponding eigenvalues such that λi > 0, for i ≤ n, and λi = 0, for i > n. Since {γ1, . . . , γk2 } is an orthonormal basis of ∑k2Hermitian matrices of Mk then A = i=1 γi ⊗ GA(γi). Now, use lemma 3.11 to obtain the required Hermitian Schmidt decompositions for each item. The converse part of each item follows from definition 3.5. Lemma 3.34. Let A ∈ M2 ⊗M2 ≃ M4 and A ∈ P4. If A is SPC then A is separable. If A = S(A) or At = S(A) then A is separable and therefore PPT. Proof. Let B ∈ M2 ⊗M2 ≃ M4. Suppose B has a Hermitian Schmidt decomposition ∑m i=1 λiγi ⊗ γi with λi > 0, for every i. Thus, the subspaces of symmetric and anti-symmetric tensors in C2 ⊗C2 are left invariant by B. Since the subspace of anti-symmetric tensors in C2 ⊗C2 is generated by w = e1 ⊗e2 −e2 ⊗e1, where {e1, e2} is the canonical basis of C2, then Bw = λw. Notice that (γi ⊗γi)w = det(γi)w. Thus, λ = ∑m i=1 λi det(γi). If A is SPC then A has a Hermitian Schmidt decomposition ∑m i=1 λiγi ⊗ γi with λi > 0, for every i, by item a) of proposition 3.33. Moreover, A is positive semidefinite. Thus, Aw = λw and λ ≥ 0. By lemma 3.32, A is separable. Now, if S(A) = A or At then S(A) is a positive semidefinite Hermitian matrix. Therefore, A has a Hermitian Schmidt decomposition ∑n i=1 αiγi ⊗ γi t with αi > 0, for every i, by item b) of proposition 3.33. Thus, At2 satisfies the same conditions of B and At2 w = λw. For these two cases, let us prove that λ = 0. By lemma 3.32, At2 is separable and A is separable. First, suppose A = S(A). Thus, At2 = S(AT )T = S(A)t2 T = At2 T , by items 5 and 7 in lemma 3.10. Since T is the flip operator and w is an anti-symmetric tensor then Tw = −w. Therefore At2 w = At2 Tw = −At2 w and At2 w = 0. Therefore λ = 0. Second, suppose At = S(A). Thus, At2 = S(AT )T = S(A)t2 T = (At)t2 T = (At2 )tT , by items 5 and 7 in lemma 3.10. Since At2 is hermitian, (At2 )t = At2 and At2 = At2 T . Since w = w and λ ∈ R then At2 w = At2 w = λw. Thus, λw = At2 w = At2 Tw = −At2 w = −λw and λ = 0. 40 Remark 3.35. In the proof of the previous theorem We saw that, if A ∈ P4 and S(A) is equal to A or At, then At2 has non null kernel. Remember that if A ∈ P4 is not PPT then At2 has full rank and has only one negative eigenvalue (see proposition 1 in [3]). Thus, A ∈ M2 ⊗M2 must be PPT and separable by Horodecki’s theorem (see [29]). However, this argument does not work for SPC matrices in M2 ⊗M2. Examples 3.36. Counterexamples for lemma 3.34 in Mk ⊗Mk, k ≥ 3: (1) The matrix C = �mq �Id ⊗ Id +D ⊗D + (iA)⊗ (iA) ∈ M3 ⊗M3 of proposition 25 in [13] is SPC, but it is not PPT. (2) As discussed in example 3.9, Id⊗Id+uut −T ∈ Mk ⊗Mk is invariant under realignment. Since its partial tranposition is Id ⊗ Id + T − uut and (Id ⊗ Id + T − uut)u = (2 − k)u then it is not PPT for k ≥ 3. Notice also that S((Id ⊗ Id + uut − T )t2 ) = S(Id ⊗ Id + T −uut) = uut +T − Id ⊗ Id and any anti-symmetric vector of Ck ⊗Ck is an eigenvector of uut + T − Id ⊗ Id associated to −2. Thus, S((Id ⊗ Id + uut − T )t2 ) is not positive semidefinite and Id ⊗ Id + uut − T is not SPC, by definiton 3.6. (3) Let A = vv t + S(vvt) ∈ Mk ⊗Mk, k ≥ 3, as in lemma 3.31. Notice that, by properties 2 and 3 in lemma 3.10 and since V = F −1(v) is Hermitian (definition 3.3), S(A) = V ⊗ V +vvt = (V ⊗V +vv t)t = (S(vvt)+vv t)t = At. Now, by lemma 3.31, FA ○GA :Mk →Mk is not completely reducible then A is not PPT or SPC, by theorems 3.2, 3.12. 3.3.2 A Remark on the Application to the Separability Problem All the results within this subsetion were published in [12]. A very useful tool to study separability in Mk ⊗Mm is the so-called filter normal form (section IV.D of [23]): If A ∈ Mk ⊗Mm ≃ Mkm is a positive definite Hermitian matrix then there exist invertible matrices R ∈ Mk and S ∈ Mm such that (R ⊗ S)A(R ∗ ⊗ S ∗) has the 1following Hermitian Schmidt decomposition: ∑n i=1 λiγi ⊗ δi, where γ1 = √ Id and δ1 = √1 m Id. k The known proof of the existence of this canonical form depends heavily on the positive definiteness of A ([34, 46]). Besides the positive definite case, there is another case where this filter normal form can be used. Assume A is PPT. By theorem 3.18, A = ∑i s =1(Vi ⊗Wi)A(Vi ⊗Wi), where ViVj = 0 and WiWj = 0, for i ≠ j. Notice that if s > 1 then A(V1 ⊗W2) = 0. Hence, A is not positive definite and we can not garantee the existence of the filter normal form for A. Now, if rank(Vi) = ki and rank(Wi) = mi then we can embed (Vi ⊗Wi)A(Vi ⊗Wi) in Mki ⊗Mmi . If (Vi ⊗Wi)A(Vi ⊗Wi) has rank kimi then its embedding in Mki ⊗Mmi is positive definite and we can obtain its filter normal form. So in this particular case, where rank 41 (Vi ⊗Wi)A(Vi ⊗Wi) is kimi for every i, the filter normal form can still be used to study the separability of (Vi⊗Wi)A(Vi⊗Wi). Recall that A is separable if and only if (Vi⊗Wi)A(Vi⊗Wi) is separable for every i (corollary 3.20). In theorem 3.18 it was shown that each (Vi ⊗Wi)A(Vi ⊗Wi) is weakly irreducible. Thus, if we could prove the existence of the filter normal form for weakly irreducible PPT matrices then this canonical form would be useful to study separability of any PPT matrix. Recall that in order to be separable a matrix must be PPT. We can obtain some inequalities for weakly irreducible PPT matrices that imply sepa­ rability, even without the filter normal form. These inequalities are based on the fact that every positive semidefinite Hermitian matrix with tensor rank 2 is separable (see theorem 3.44). We want to emphasize that the filter normal form would also be useful to sharpen these inequalities (see example 3.38). If A ∈ Mk ⊗Mm is PPT or SPC or invariant under realignment then FA ○GA :Mk →Mk is completely reducible. By corollary 3.20, the Separability Problem is reduced to the weakly irreducible case. Let A ∈ Mk ⊗Mm ≃ Mkm be a weakly irreducible PPT or SPC or invariant under realign­ ment matrix. By proposition 3.17, A has the following Hermitian Schmidt decomposition: ∑n i=1 λiγi ⊗ δi, such that λ1 > λ2 ≥ . . . ≥ λn > 0 and I(γi) ⊂ I(γ1), I(δi) ⊂ I(δ1), for 1 ≤ i ≤ n. Proposition 3.37. Let A ∈ Mk ⊗Mm ≃ Mkm and A ∈ Pkm. Let A be a weakly irreducible PPT or SPC or matrix invariant under realignment. Let ∑n i=1 λiγi ⊗ δi be a Hermitian Schmidt decomposition of A such that λ1 > λ2 ≥ . . . ≥ λn > 0 and I(γi) ⊂ I(γ1), I(δi) ⊂ I(δ1), for 1 ≤ i ≤ n, by proposition 3.17. Let µ be the smallest positive eigenvalue of γ1 ⊗ δ1. λ1µ(1) If λ2+...+λn ≥ 1 then A is separable. λ1µ ≥ 1(2) If A is SPC or invariant under realignment and λ2+...+λn 2 then A is separable. Proof. (1) Notice that FA ○GA(γi) = λi 2γi for 1 ≤ i ≤ n. Thus, the eigenvalues of FA ○GA : Mk → Mk are λi 2 , 1 ≤ i ≤ n, and possibly 0. Hence, the largest eigenvalue of FA ○GA is λ1 2 . Since FA ○ GA : Mk → Mk is a self-adjoint positive map then associated to this eigenvalue there exists γ ∈ Pk such that FA ○GA(γ) = λ1 2γ, by theorem 2.8 (or [21, Proposition 2.5]). Since λ2 > λ2 > 0, for every 1 ≤ i ≤ n, then the multiplicity of λ2 is 1. Therefore,1 i 1 γ1 = λγ, for some λ ∈ R, since γ1 and γ are Hermitian. Hence, δ1 = GA(γ1) = λGA(γ) and γ1 ⊗ δ1 = λ2γ ⊗ GA(γ) is a positive semidefinite Hermitian matrix, since γ ∈ Pk and GA :Mk →Mm is a positive map. Notice that the smallest positive eigenvalue of µ 1 γ1 ⊗ δ1 is 1 and, since tr(γi 2) = tr(δ2) = 1, the smallest eigenvalue of γi ⊗δi is greater or equal to -1. By hypothesis, I(γi ⊗δi) ⊂ I(γ i 1 ⊗δ1) 42 then µ 1 γ1 ⊗ δ1 + γi ⊗ δi is positive semidefinite and, by theorem 3.44, it is separable. Now if λ1µ ≥ λ2 + ... + λn then A = (λ1µ −∑i n =2 λi)(µ 1 γ1 ⊗ δ1) +∑i n =2 λi(µ 1 γ1 ⊗ δ1 + γi ⊗ δi). Notice that all the matrices inside parentheses are separable. (2) Since FA ○GA(γi) = λi 2γi, for 1 ≤ i ≤ n, then by lemma 3.11, δi = GA(γi) = γi, if A is SPC, and δi = GA(γi) = γit, if A is invariant under realignment. In any case, since tr(γi 2) = 1 then the smallest eigenvalue of γi ⊗ δi (δi = γi or γit) is greater or equal to −2 1 . λiFinally, repeat the argument of item (1) and write A = (λ1µ − ∑n i=2 2 )(µ 1 γ1 ⊗ δ1) + ∑n i=2 λi(2 1 µ γ1 ⊗ δ1 + γi ⊗ δi). Note that if λ1µ ≥ 2 1 (λ2 + ... + λn) then all the matrices inside parentheses are separable by theorem 3.44. Example 3.38. Let {γ1, γ2, γ3, γ4} be the normalized Pauli’s basis of M2, where γ1 = √1 Id. It is known that a necessary and sufficient condition for the separability of ∑4 i=1 λiγi ⊗ 2 γi, λi ≥ 0, is the inequality of item (2) (see [34]). Tensor Rank 2 Implies Separability in Mk1 ⊗ . . . ⊗Mkn Here, we show that tensor rank 2 in Mk1 ⊗. . .⊗Mkn implies separability of positive semidefinite Hermitian matrices in Mk1 ⊗ . . . ⊗Mkn = Mk1 ⊗ (Mk2 ⊗ . . . ⊗Mkn ) ≃ Mk1...kn . First, let us recall some definitions and some well known results regarding tensor rank. Definition 3.39. Let V1 ⊗ . . . ⊗ Vn be the tensor product space of the complex vector spaces Vi (1 ≤ i ≤ n) over the complex field. Let r ∈ V1 ⊗ . . . ⊗ Vn. The tensor rank of r is 1, if r = v1 ⊗ . . . ⊗vn and r ≠ 0. The tensor rank of r is the minimal number of tensors with tensor rank 1 that can be added to form r. Theorem 3.40. (Marcus-Moyls [37]) Let V1 and V2 be complex vector spaces and let vi, rj ∈ V1 and wi, sj ∈ V2, for 1 ≤ i ≤ n and 1 ≤ j ≤ k. Let ∑i n =1 vi ⊗wi = ∑j k =1 rj ⊗ sj ∈ V1 ⊗ V2. (i) If {v1, ..., vn} is a linear independent set then span{w1, ..., wn} ⊂ span{s1, ..., sk}. (ii) If {w1, ..., wn} is a linear independent set then span{v1, ..., vn} ⊂ span{r1, ..., rk}. Corollary 3.41. Let ∑n i=1 vi ⊗ wi = ∑j k =1 rj ⊗ sj . If {v1, ..., vn} and {w1, ..., wn} are linear independent sets then k ≥ n. So the tensor rank of ∑i n =1 vi ⊗wi is n. Recall that Mk stands for the set of complex matrices of order k and Pk for the subset of positive semidefinite Hermitian matrices of Mk. We are also identifying Mk1 ⊗ . . . ⊗Mkn with Mk1...kn via Kronecker product. Lemma 3.42. Let A ∈ Mk ⊗ Mm, A ∈ Pkm and tensor rank(A) = n. We can write A = ∑n i=1 γi ⊗ δi, where γi ∈ Mk, δi ∈ Mm are Hermitian matrices such that I(γi) ⊂ I(γ1) and I(δi) ⊂ I(δ1), for every i, and γ1 ∈ Pk, δ1 ∈ Pm. 43 http:Theorem3.40 Proof. Since A ∈ Mk ⊗Mm is Hermitian with tensor rank n, we can write A = ∑n i=1 Ai ⊗Bi, where {A1, . . . ,An} and {B1, . . . ,Bn} are linear independent sets of Hermitian matrices (see = ∑ n i=1 tr(Bi)Aiminimal Hermitian decomposition in [12] or corollary 3.41). Let γ1 and δ1 = tr(A) ∑n i=1 tr(Ai)Bi. Since A ∈ Pkm then FA :Mm →Mk and GA :Mk →Mm are positive maps then ( Id FA trA ) = γ1 ∈ Pk and GA(Id) = δ1 ∈ Pm. First, let us prove that B = A − γ1 ⊗ δ1 has tensor rank n − 1. Notice that B is Hermitian and let B = ∑j l =1 γ ′ ⊗δ ′ , where {γ ′ , . . . , γ ′} and {δ ′ , . . . , δ ′} arej j 1 l 1 l linear independent sets of Hermitian matrices. Notice that l ≥ n − 1, otherwise A would have tensor rank smaller than n. Now, since {γ1′ , . . . , γ l ′} is a linear independent set, by theorem 3.40, we have span{δ1′ , . . . , δ l ′} ⊂ span{B1, . . . ,Bn, δ1} = span{B1, . . . ,Bn}. Thus, l ≤ n. Next, consider the trace inner product. Let δ ′ be the projection of the Id inside the span{B1, . . . ,Bn}. Thus, tr(Biδ ′) = tr(BiId) = tr(Bi) for every 1 ≤ i ≤ n. Notice that δ ′ ≠ 0, otherwise tr(Bi) = 0 for every 1 ≤ i ≤ n and tr(A) = 0. Now, 0 = ∑i n =1 Aitr(Bi)− γ1tr(δ1) = ∑i n =1 Aitr(Biδ ′)− γ1tr(δ1δ ′) = FB (δ ′) = ∑j l =1 γj ′ tr(δj ′ δ ′). Since {γ1′ , . . . , γ l ′} is a linear independent set then we get tr(δj ′ δ ′) = 0 for 1 ≤ j ≤ n. Since δ ′ ∈ span{B1, . . . ,Bn} and δ ′ is orthogonal to span{δ1′ , . . . , δ l ′} ⊂ span{B1, . . . ,Bn} then l ≤ n−1. Hence, l = n − 1 and B has tensor rank n − 1, by corollary 3.41. Thus, let us write A = ∑i n =1 γi ⊗ δi, such that {γ1, . . . , γn} and {δ1, . . . , δn} are linear independent sets of Hermitian matrices and γ1, δ1 as defined above. Since {γ1, . . . , γn} is a linear independent set of Hermitian matrices then there exist Hermitian matrices C1, . . . ,Cn such that tr(γiCj ) = δij . For each Ci there exists qi > 0 such that Id ± qiCi ∈ Pk. Since A ∈ Pkm then GA : Mk → Mm is a positive map and GA(Id ± qiCi) = δ1 ± qiδi ∈ Pm. By lemma 2.1, I(δi) ⊂ I(δ1), for every i. Analogously, we obtain I(γi) ⊂ I(γ1), for every i. Remark 3.43. Notice that γ1 and δ1 in the proof of lemma 3.42 are multiples of the so-called marginal states of A (see [30]). Theorem 3.44. Let A ∈ Mk1 ⊗ . . . ⊗Mkn and A ∈ Pk1...kn . If A has tensor rank smaller or equal to 2 then A is separable. Proof. Let A = A1 ⊗ . . . ⊗ An + B1 ⊗ . . . ⊗ Bn. Thus, A as an element of Mk1 ⊗Mk2...kn has tensor rank smaller or equal to 2. If A has tensor rank 1 in Mk1 ⊗Mk2...kn then A = γ1 ⊗ δ1, where γ1 ∈ Pk1 , δ1 ∈ Pk2...kn . By theorem 3.40, δ1 ∈ span{A2 ⊗ . . .⊗An,B2 ⊗ . . .⊗Bn}. So δ1 has tensor rank smaller or equal to 2 in Mk2 ⊗ . . . ⊗Mkn and by induction on n, δ1 is separable in Mk2 ⊗ . . . ⊗Mkn . Therefore, A is separable in Mk1 ⊗ . . . ⊗Mkn . 44 Now, assume A has tensor rank 2 in Mk1 ⊗Mk2...kn . By lemma 3.42, A = γ1 ⊗ δ1 + γ2 ⊗ δ2 such that γ1 ∈ Pk1 , δ1 ∈ Pk2...kn and γ2 ∈ Mk1 , δ2 ∈ Mk2...kn are Hermitian matrices such that I(γ2) ⊂ I(γ1), I(δ2) ⊂ I(δ1). Choose 0 ≠ λ ∈ R such that γ1 − λγ2 ∈ Pk and 0 ≠ v ∈ ker(γ1 − λγ2) ∩ I(γ1), by lemma 2.2. Notice that A = (γ1 − λγ2)⊗ δ1 + γ2 ⊗ (δ2 + λδ1). Since A ∈ Pk1...kn then GA : Mk1 →Mk2...km is a positive map. Since tr((γ1 − λγ2)vv t) = 0 t) tr(γ1vv t)then GA(vv = tr(γ2vv t)(δ2 + λδ1) ∈ Pm. Notice that 0 ≠ λ = tr(γ2vv t), since v ∈ I(γ1) and γ1 ∈ Pk. γ2Now, let β1 = δ1, β2 = tr(γ2vv t)(δ2 +λδ1), α1 = γ1 −λγ2, α2 = tr(γ2vvt) . Notice that α1, β2, β1 are positive semidefinite Hermitian matrices such that I(β2) ⊂ I(β1) and A = α1 ⊗β1 +α2 ⊗β2. Next choose 0 ≠ q ∈ R such that β1 − qβ2 is positive semidefinite and 0 ≠ w ∈ ker(β1 − qβ2)∩ I(β1), by lemma 2.2. Notice that A = α1 ⊗ (β1 − qβ2) + (α2 + qα1)⊗ β2. Since A ∈ Pk1...kn then FA : Mk2...km →Mk1 is a positive map. Since tr((β1 − qβ2)ww t) = 0 then FA(ww t) = tr(β2ww t)(α2 + qα1) ∈ Pk. Notice also that 0 ≠ tr(β1 � ww t) = tr(β2ww t), since β1 ∈ Pm and w ∈ I(β1). Since tr(β2ww t) > 0, by the positive semidefiniteness of β2, we obtain the following β2minimal separable decomposition: A = α1 ⊗ (β1 − qβ2) + tr(β2ww t)(α2 + qα1)⊗ tr(β2ww t) . Now since α1 and tr(β2ww t)(α2 + qα1) are linear independent, because A has tensor rank β22 in Mk1 ⊗Mk2...kn , then (β1 −qβ2) and tr(β2ww t) belong to the span{A2 ⊗. . .⊗An,B2 ⊗. . .⊗Bn}, β2by theorem 3.40. Thus (β1 − qβ2) and tr(β2wwt) are positive semidefinite Hermitian matrices with tensor rank smaller or equal to 2 in Mk2 ⊗. . .⊗Mkn and, by induction on n, (β1 −qβ2) and β2 tr(β2wwt) are separable in Mk2 ⊗ . . . ⊗Mkn . Therefore, A is separable in Mk1 ⊗ . . . ⊗Mkn . Remark 3.45. There is a generalization of this result in M2 ⊗Mm. Every A ∈ P2m ⊂ M2 ⊗Mm with tensor rank 3 is separable (See theorem 19 in [13]). However, this is not true in M3 ⊗M3 (See proposition 25 in [13]). 45 46 Chapter 4 An application of Borsuk-Ulam Theorem All the results within this chapter were published in [11]. Given a topological space D, let us denote by C�(D) the subset of the vector space C (D) of all real-valued continuous functions on D formed by the functions that attain the maximum exactly once in D. The set C�(D) fails to be a vector space for many reasons, for example the zero function does not belong to C�(D). Gurariy and Quarta asked the following question: Is it possible to find a linear subspace V of C (D) such that V ⊂ C�(D) ∪ {0}? If so, how big can be the dimension of V ? The main results obtained by Gurariy and Quarta in this direction are the following: (A) There is a 2-dimensional linear subspace of C[a, b) contained in C�[a, b) ∪ {0}. (B) There is a 2-dimensional linear subspace of C(R) contained in C�(R) ∪ {0}. (C) There is no 2-dimensional linear subspace of C [a, b] contained in C� [a, b] ∪ {0}. The purpose of this chapter is to obtain far-reaching generalizations of the aforementioned results of Gurariy and Quarta. We investigate the existence of n-dimensional subspaces – instead of 2-dimensional subspaces – formed by functions that attain the maximum exactly once (question posed in [9, Problem 2.9]). While Gurariy and Quarta [27] used typical analyt­ ical techniques, the manifested nature of the problem led us to apply topological techniques, for example the Borsuk-Ulam theorem. 47 Our main result is theorem 4.1: If D is a compact subset of Rm and V is a linear subspace of C(D) such that V ⊂ C�(D) ∪ {0} then dim(V ) ≤ m. Thus, we recover theorem (C) of Gurariy and Quarta. Moreover, this inequality is sharp. Gurariy and Quarta also asked if there exists a 3-dimensional linear subspace of C[a, b) contained in C�[a, b)∪ {0}. We can not prove or disprove the existence of this subspace. Our approach seems to be useless when we replace the hypothesis of compactness of D ⊂ Rm by σ−compactness. However, it might be possible to use Borsuk-Ulam theorem locally and some other topological features in order to tackle this problem. We shall describe in the final section of this chapter an approach that seems promissing and one open question. An affirmative answer to this question would imply a complete solution for the problem. 4.1 Main Result Theorem 4.1. If D is a compact subset of Rn and V is a linear subspace of C(D) such that V ⊂ C�(D) ∪ {0} then dim(V ) ≤ n. Proof. Let f1(x), . . . , fm(x) be a basis of V . Thus, any non null linear combination of these functions attains it maximum exactly once in K. Define F : D → Rm, F (x) = (f1(x), . . . , fm(x)). Notice that F is continuous since every fi is continuous. Notice that every function of V can be written as ⟨v, F (x)⟩, where v ∈ Rm. Define f : Sm−1 →D as f(v) = the unique point of maximum of ⟨v, F (x)⟩ in D. n→∞By contradiction assume that f is not continuous. Thus, there is a sequence vn �→ v and ) j→∞�f(vn)−f(v)� > q. Since f(vn) ∈ D and D is compact, there is a subsequence f(vnj �→ y ∈ D. By definition of f(vnj ), we have ⟨vnj , F (f(vnj ))⟩ ≥ ⟨vnj , F (x)⟩, for every x ∈ D. If we fix x and let j → ∞, since F is continuous, we get ⟨v, F (y)⟩ ≥ ⟨v, F (x)⟩, for every x ∈ D. ) j→∞Thus, y = f(v) and f(vnj �→ f(v) ∈ D, but �f(vnj ) − f(v)� > q. This is a contradiction. So f : Sm−1 →D is continuous. Remind that D ⊂ Rn. Finally, if m > n then by Borsuk-Ulam theorem (see, for example, [19]) there is a pair of antipodal points s and −s in Sm−1 such that f(s) = f(−s). Hence, the point of maximum of ⟨s, F (x)⟩ in D is the point of maximum of ⟨−s, F (x)⟩ in D, which is the point of minimum of ⟨s, F (x)⟩ in D. Thus, ⟨s, F (x)⟩ is constant and does not belong to C�(D), which is a contradiction. Remark 4.2. Since the Euclidean sphere Sm−1 ⊂ Rm is compact and every linear functional defined on Rm restricted to Sm−1 attains its maximum at only one point of Sm−1 then there is a m−dimensional subspace V of C(Sm−1) such that V ⊂ C�(Sm−1) ∪ {0}. Thus, the upper bound for the dimension of the vector space V in the previous theorem is sharp. 48 4.2 An Infinite Dimensional Example Let D be a compact subset of Rn. We saw above that, for n < m, there is no m-dimensional subspace of C(D) formed, up to the origin, by functions that attain the maximum only at one point. In this section we show that if we allow D to be a compact subset of an infinite dimensional Banach space, C�(D) may contain, up to the origin, an infinite dimensional subspace of C(D). Example 4.3. Let D be the following subset of �2: ∞ )∞ )∞ D = {(an ) : (an n=1 ∈ �2 and �(an n=1�2 ≤ 1} . n n=1 1It is clear that D is a subset of the Hilbert cube ∏∞ [− 1 , ]. Since the Hilbert cube is n=1 n n compact, to prove that D is compact it is enough to show that it is closed. Let (vj j=1 =)∞ ) ∞ ∞ )∞ n=1((v n j n ) be a sequence in D converging to w = (wn ∈ �2. Since convergence in �2 n=1 j=1 j nv nj j nimplies coordinatewise convergence, wn , for every fixed n. For every = = lim lim , so nwn v j k, k k k j n)∞ n=1�22 ≤ 1.�wn� j�v � j�v �2 ≤ lim sup �(vn∑ n 2 2 =∑ ∑ 2 =lim lim n j j jn=1 n=1 n=1 This shows that �(nwn)∞ n=1�2 ≤ 1, proving that w ∈ D. So D is a compact subset of �2. Now we proceed to show that C�(D) ∪ {0} contains an infinite dimensional subspace of C(D). Consider the function F :D �→ �2 , F ((an ) ∞ ) = (an)∞ n=1 . n n=1 Let b = (c1, c2, . . .) ∈ �2, bn = (c1, . . . , cn, 0, 0, . . .) and φb: �2 �→ R, φb(x) = ⟨b, x⟩. Consider φb ○F :D �→ R and note that �φbn ○F (x)−φb ○F (x)� ≤ �bn −b�2�F (x)�2 ≤ �bn −b�2, since �F (x)�2 ≤ 1. Thus, φb ○ F : D �→ R is continuous as a uniform limit of a sequence of continuous functions (φbn ○ F (x))∞ n=1. b bNext, φb ○ F (x) = ⟨b, F (x)⟩ < ⟨b, ⟩ whenever F (x) ≠ . As F is a bijection onto the �b�2 �b�2 closed unit ball of �2, there is a unique y ∈ D such that F (y) = . This shows that φb ○F (x) attains its maximum only at y. Finally {φb ○ F : D �→ R, b ∈ �b � b � 2 2 } is a c−dimensional vector space. 49 4.3 Open Problem Gurariy and Quarta asked if there exists a 3-dimensional linear subspace of C[a, b) contained in C�[a, b) ∪ {0}. Here, we shall describe an approach that might be useful to disprove the existence of such subspace. Assume that there is a 3 dimensional subspace of C[a, b) contained in C�[a, b) ∪ {0}. Let f1(x), f2(x), f3(x) be a basis. Define f : S2 → [a, b) as f(v) = the unique point of maximum of ⟨v, F (x)⟩ in [a, b), where F (x) = (f1(x), f2(x), f3(x)). Let [a, b) = ⋃ [a, b− 1 ] for some nn>k,n∈N suitable k ∈ N. 1 1 1Notice that f−1([a, b − ]) is closed in S2 and f : f−1([a, b − ])→ [a, b − ] is continuous, n n n by the same argument that was used in the proof of theorem 4.1. 1 1Now, for every c ∈ [a, b − n ], we have f−1(c) ⊂ f−1([a, b − n ]) and there is n such that ∞ 1int(f−1([a, b − n ])) ≠ ∅ in S2, since S2 = ⋃ f−1([a, b − 1 ]), by Baire category theorem. nn>k,n∈N 1 1But can we find n ∈ N and c ∈ [a, b − n ] such that ∂(f−1(c)) ⊂ int(f−1([a, b − n ]))? Let us assume that the following conjecture is true and let us obtain a contradiction. Thus, the key result to disprove the existence of this subspace is the following conjecture. 1 1Conjecture 4.4. There is n ∈ N and c ∈ [a, b − n ] such that ∂(f−1(c)) ⊂ int(f−1([a, b − n ])), where ∂(A) and int(A) mean the frontier and the interior of A in S2, respectively. In order to obtain a contradiction, we need the following lemma : Lemma 4.5. If x1, x2 ∈ f−1(c) ⊂ S2 then the geodesic arc that connects x1 and x2, which shall be denoted by x1x2, is also contained in f−1(c). Proof. Notice that x2 ≠ −x1, otherwise c would be the point of maximum of ⟨x1, F (x)⟩ and ⟨x2, F (x)⟩ = −⟨x1, F (x)⟩. Thus, ⟨x1, F (x)⟩ would be constant and ⟨x1, F (x)⟩ ∉ C�[a, b) ∪ 0. The geodesic arc connecting x1 to x2 is (1−t)x1+tx2 , 0 ≤ t ≤ 1.l(1−t)x1+tx2l 1−t tNotice that ⟨ (1−t)x1+tx2 , F (x)⟩ = ⟨x1, F (x)⟩ + ⟨x2, F (x)⟩l(1−t)x1+tx2l l(1−t)x1+tx2l l(1−t)x1+tx2l 1−t t ⟨ (1−t)x1+tx2≤ l(1−t)x1+tx2l ⟨x1, F (c)⟩ + l(1−t)x1+tx2l ⟨x2, F (c)⟩ = l(1−t)x1+tx2l , F (c)⟩, for every x ∈ [a, b). Thus, f( (1−t)x1+tx2 ) = c for every 0 ≤ t ≤ 1.l(1−t)x1+tx2l 50 1 1Next, let x0 ∈ ∂(f−1(c)) ⊂ int(f−1([a, b − n ])). There is a small circle S ⊂ int(f−1([a, b − n ])) around x0. 1The function f : S → [a, b − n ] is continuous, by Borsuk-Ulam theorem, there is a pair of antipodal points in S, x1 and x2, such that f(x1) = f(x2) = d. By the previous lemma, the geodesic arc x1x2 ⊂ f−1(d). Since x0 ∈ x1x2 then c = d, because f−1(c) is closed and x0 ∈ ∂(f−1(c)) ⊂ f−1(c). Thus, x1x2 ⊂ f−1(c) (see figure 1 below). If i ∈ x1x2 ∩ int(f−1(c)) then x0 ∈ int(f−1(c)) (We can connect, by geodesic arcs, all the points of a neighborhood of i within int(f−1(c)) to x1 and x2. Thus, x0 ∈ int(f−1(c)). See figure 2 below). So x1x2 ⊂ ∂(f−1(c)). Now, consider the circle S1, centered in C = (0, 0, 0), which contains the geodesic arc x1x2. Notice that if z ∈ ∂(f−1(c)) ∩ S1 then we can repeat the argument and obtain a geodesic arc z1z2 ⊂ ∂(f−1(c)) such that z ∈ z1z2 (see figure 3 below). If z1z2 is not contained in S1 then we can prove that x1 or x2 belongs to int(f−1(c)) (by connecting the points of z1z2 and x1x2 via geodesic arcs), which is a contradiction. Thus, z1z2 ⊂ ∂(f−1(c)) ∩ S1. We have just proved that ∂(f−1(c))∩S1 is open in S1, but it is also closed as an intersection of closed sets. Since S1 is connected then ∂(f−1(c)) ∩ S1 = S1. Thus, there are antipodal points in S1, s and −s, such that f(s) = f(−s) = c, but this is a contradiction. 51 52 Chapter 5 Basic Sequences in �p spaces All the results of this chapter were published in [15]. During a Non-linear Analysis Seminar at Kent State University (Kent, Ohio, USA) in 2003, Richard M. Aron and Vladimir I. Gurariy posed the following question: Question 5.1 (R. Aron & V. Gurariy, 2003). Is there an infinite dimensional closed subspace of �∞ every nonzero element of which has only a finite number of zero coordinates? Question 5.1 has also appeared in several recent works (see, e.g., [9, 20, 22, 38]) and, for the last decade, there have been several attempts to partially answer it, although nothing conclusive in relation to the original problem has been obtained so far. Throughout this chapter, and if X denotes a sequence space, we shall denote by Z(X) the subset of X formed by sequences having only a finite number of zero coordinates. Here, we shall provide (among other results) the definitive answer to Question 5.1. Namely, if X stands for c0, or �p, with p ∈ [1, ∞], we prove the following: (i) There is no infinite dimensional closed subspace of X inside Z(X)∪{0}(Corollaries 5.7 and 5.16). (ii) There exists an infinite dimensional closed subspace of X inside V ∖ Z(V ) ∪ {0}, for any infinite dimensional closed subspace V of X (Theorem 5.18). In order to obtain the results above we shall make use of Functional Analysis techniques, basic sequences, complemented subspaces, and some classical Linear Algebra and Real Anal­ ysis approaches. From now on, if Y is any sequence space and y ∈ Y , then y(j) shall denote the j−th coordinate of y with respect to the canonical basis (ej )j . Also, if (mk)k∈N is a subse­ quence of (nk)k∈N, we shall write (mk)k∈N ⊂ (nk)k∈N. If V is a normed space and (vk)k∈N ⊂ V , we denote by ⟨v1, v2, . . .⟩ the linear span of {v1, v2, . . . } and by [v1, v2, . . . ] the closed linear span of {v1, v2, . . . }. If W ⊂ V , we denote S1(W ) = {w ∈ W, �w� = 1}. The rest of the notation shall be rather usual. 53 � � � 5.1 The case: X = �p, p ∈ [1, ∞[ We need a series of technical lemmas in order to achieve the main result of this section. We believe that these lemmas are of independent interest. Lemma 5.2. Let V be an infinite dimensional closed subspace of �p, p ∈ [1, ∞[. Given 0 < q < 4 there is an increasing sequence of natural numbers (sk)k∈N and a normalized basic sequence 33 (fk)k∈N ⊂ V such that (1) fk(sj ) = 0 for 1 ≤ j ≤ k − 1. (2) f1(s1) ≠ 0. (3) �f1(sk+1)� + . . . + �fk(sk+1)� < 2k+1 �fk+1(sk+1)� for every k (thus fk(sk) ≠ 0 for every k ∈ N). (4) (fk)k∈N has basis constant smaller than 84 − − 2 9 � � . (5) [f1, f2, . . . ] is complemented in �p with a projection Q : �p → �p of norm ��Q�� ≤ 8−2� .4−33� Proof. Let f1 ∈ V be such that �f1�p = 1. Let N1 ∈ N be such that (1) f1(N1) ≠ 0. (12 �(f1(n))n∞=N1+1�p < 2 � 2 . Let s1 = N1. Suppose we have defined f2, . . . , ft ∈ V and s1 = N1 < s2 < N2 < . . . < st < Nt such that (1) �fk�p = 1 for 1 < k ≤ t (2) fk(n) = 0 for 1 ≤ n ≤ Nk−1 for every 1 < k ≤ t (Thus fk(sj ) = 0 for 1 ≤ j ≤ k − 1 since sk−1 < Nk−1). (3) �(�f1(n)� + . . . + �fk(n)�)∞ n=Nk+1�p < 2k+1 for 1 < k ≤ t (4) �f1(sk+1)� + . . . + �fk(sk+1)� < 2k+1 �fk+1(sk+1)� for 1 < k ≤ t − 1. (Thus fk(sk) ≠ 0 for 1 < k ≤ t) 54 � Since V is an infinite dimensional closed subspace of �p, there exists ft+1 ∈ V such that �ft+1�p = 1 and ft+1(1) = . . . = ft+1(Nt) = 0. Now, if there is no n > Nt such that �f1(n)� + . . . + �ft(n)� < 2t+1 �ft+1(n)� then q q q > �(�f1(n)� + . . . + �ft(n)�)∞ n=Nt+1�p ≥ �ft+1�p = , 2t+1 2t+1 2t+1 which is absurd. Therefore there exist st+1 > Nt such that q �f1(st+1)� + . . . + �ft(st+1)� < �ft+1(st+1)�. 2t+1 Next, since (�f1(n)� + . . . + �ft+1(n)�)n∈N ∈ �p then there exist Nt+1 > st+1 such that q �(�f1(n)� + . . . + �ft+1(n)�)∞ n=Nt+1+1�p < 2t+2 . The induction to construct (fk)k∈N enjoying the four properties above is now complete. Now, in order to show that (fk)k∈N is a basic sequence, let us define (n), if 1 ≤ n ≤ N1 (n), if Nk−1 < n ≤ Nkf̃1(n) = { f1 f̃k(n) = { fk 0, otherwise 0, otherwise Notice that f̃k ≠ 0, since Nk−1 < sk < Nk and f̃k(sk) = fk(sk) ≠ 0. Note also that (f̃k)k∈N is a block basis of the canonical basis of �p. Since q�(�f1(n)� + . . . + �fk(n)�)∞ n=Nk+1�p < 2k+1 , then q�(fk(n))n∞=Nk+1�p < 2k+1 . Now since fk(n) = 0 for 1 ≤ n ≤ Nk−1 we obtain q q 1 − ≤ �f̃k�p ≤ 1 and �fk − f̃k�p < 2k+1 2k+1 ̃� fkfor k ∈ N. In particular, 44 −� = 1 − 4 ≤ �f̃k�p ≤ 1 for every k ∈ N. Let gk = lf̃k lp for every k. Notice ∑∞that (gk)k∈N is a normalized block basis of the canonical basis of �p. So � k=1 akgk�p = �(ak)k∈N�p and (gk)k∈N has basis constant K = 1. Let {σk, k ∈ N} be the following partition of N: σ1 = {1, . . . ,N1} and σk = {Nk−1 + 1, . . . ,Nk}. 55 � � Next, let Ek = {f ∈ �p, f(i) = 0, for i ∉ σk}. Thus, gk ∈ Ek and by [32, Theorem 30.18] the closed subspace [g1, g2, . . .] is complemented in �p with a projection P : �p → �p of norm 1. Let us now prove that (fk)k∈N is equivalent to (gk)k∈N and [f1, f2, . . .] is also complemented in �p. Indeed, ̃ ̃fk − fk fk fk�fk − gk�p = �fk − �p ≤ �fk �p + � − �p �f̃kp� �f̃k�p �f̃k�p �f̃k�p ≤ 1 − �f̃k�p 1 q 4 ̃ q+ ≤ (1 − �fk�p + ) �f̃k�p �f̃k�p 2 k+1 4 − q 2k+1 4 ( 2q≤ ) . 4 − q 2k+1 Thus, (gk)k∈N is a normalized basic sequence such that [g1, g2, . . .] is complemented in �p with a projection P : �p → �p of norm 1 and ∞ �fk − gk�p ≤ ∞ ∑ 4 q 4q=∑ δ := 4 − q . 4 − q2k k=1 k=1 Since 0 < q < 4 , we obtain 8Kδ��P �� = 8δ ≤ 8 4� < 1. By the principle of small perturba­33 4−� tion ([16, Theorem 4.5]) the sequence (fk)k∈N is equivalent to (gk)k∈N and [f1, f2, . . .] is also complemented in �p. Finally, let us compute an upper bound for the basis constant of (fk)k∈N and for the norm of the projection Q : �p → �p onto [f1, f2, . . .]. First, the linear transformation T (∑∞ k=1 akgk) = k=1 akfk is an invertible continuous linear ∑∞ transformation from the closed span of (gk)k∈N to the closed span of (fk)k∈N. In the proof [16, Theorem 4.5] it is shown that ��T �� ≤ (1 + 2Kδ) ≤ (1 + 8δ) ≤ 2 and ��T −1�� ≤ (1 − 2Kδ)−1. Let Pn(∑∞ k=1 akgk) = ∑k n =1 akgk. Notice that ��Pn�� = 1. Thus, for n ≤ m, n m m −1( −1����p = �T ○ Pn ○ T )�p ≤ ��T �� ��Pn�� ��T∑ akfk ∑ akfk ∑ akfk p k=1 k=1 k=1 m ≤ 2 � akfk�∑ 1 − 2Kδ p. k=1 2 ≤ 8−2�is smaller than 1−2Kδ 4−9�Then, the basis constant of (fk)k∈N 1 and δ ≤ 4� 4−� . , since K = Again, using [16, Theorem 4.5], the linear transformation Id − (T ○ P ) : [f1, f2, . . .]→ [f1, f2, . . .] is invertible and has norm smaller than 8Kδ��P �� = 8δ < 1. Therefore, there exists an inverse for S = T ○ P : [f1, f2, . . .] → [f1, f2, . . .] with norm 1��S−1�� ≤ 1−8δ . Now Q = S−1 ○ (T ○ P ) : �p → �p is a projection onto [f1, f2, . . .] with norm ��Q�� ≤ ��S−1�� ��T �� ��P �� = × 2 × 1 ≤ 8−2� , since δ ≤ 4� .1− 1 8δ 4−33� 4−� 56 � � Remark 5.3. In the previous theorem, note that the convergence of ∑∞ k=1 akfk implies the convergence ∑k ∞ =1 a2kf2k and ∑k ∞ =1 a2k−1f2k−1, since (gk)k∞=1 is a block basis of the canonical basis of �p and (fk)k∈Nis equivalent to (gk)k∈N. Lemma 5.4. Let V be an infinite dimensional closed subspace of �p, p ∈ [1, ∞[. There exist an increasing sequence of natural numbers (sk)k∈N and a basic sequence (lsk )k∈N ⊂ V such that (1) lsk (sk) ≠ 0 (2) lsk (sj ) = 0 for k ≠ j (3) [ls1 , ls2 , . . .] is complemented in �p. Proof. Let 0 < q < 1 . Then 4 − 9q > 1, 4 − 33q > 1 and512 8q (8 − 2q)( 8 − 2q ) < 512q < 1. 4 − 9q 4 − 33q Let (sk)k∈N and (fk)k∈N be as in Lemma 5.2, using this q. Define l0,k = fk. Notice that l0,k(sk) = fk(sk) ≠ 0 and l0,k(sj ) = 0 for sj ∈ {s1, . . . , sk}∖{sk}. Define − l0,k(sk+1) l1,k = l0,k ) fk+1. fk+1(sk+1 Notice that (1) l1,k(sj ) = 0 for sj ∈ {s1, . . . , sk, sk+1} ∖ {sk}. (2) l1,k(sk) = fk(sk) ≠ 0. � lfk(sk+1)l �(3) Since �l0,k(sk+1)� = �fk(sk+1)� < 2k+1 �fk+1(sk+1)� thus lfk+1(sk+1)l < 2k+1 < 1 and �l1,k(n)� ≤ �fk(n)� + �fk+1(n)� for every n ∈ N. (4) �l1,k − l0,k�p < 2k+1 �fk+1�p = 2k+1 . Suppose we have already defined l0,k, . . . , lt,k such that (1) li,k(sj ) = 0 for sj ∈ {s1, . . . , sk+i} ∖ {sk}, for 1 ≤ i ≤ t (2) li,k(sk) = fk(sk) ≠ 0, for 1 ≤ i ≤ t 57 � � � � (3) �li,k(n)� ≤ �fk(n)� + . . . + �fk+i(n)�, for every n ∈ N and for 1 ≤ i ≤ t (4) �li,k − li−1,k�p < 2k+i , for 1 ≤ i ≤ t. − lt,k(sk+t+1)Define lt+1,k = lt,k fk+t+1. Since fk+t+1(sj ) = 0 for 1 ≤ j ≤ k + t then lt+1,k(sj ) = fk+t+1(sk+t+1) lt,k(sj ) for 1 ≤ j ≤ k + t. Since lt+1,k(sk+t+1) = 0 then (1) lt+1,k(sj ) = 0 for sj ∈ {s1, . . . , sk+t+1} ∖ {sk} (2) lt+1,k(sk) = lt,k(sk) = fk(sk) ≠ 0 (3) �lt,k(sk+t+1)� ≤ �fk(sk+t+1)� + . . . + �fk+t(sk+t+1)� ≤ �f1(sk+t+1)� + . . . + �fk+t(sk+t+1)� < 2k+t+1 �fk+t+1(sk+t+1)�. llt,k(sk+t+1)l � Therefore lfk+t+1(sk+t+1)l < 2k+t+1 < 1 and �lt+1,k(n)� ≤ �lt,k(n)� + �fk+t+1(n)� ≤ �fk(n)� + . . . + �fk+t+1(n)� for every n ∈ N. (4) �lt+1,k − lt,k�p < 2k+t+1 �fk+t+1�p = 2k+t+1 . The induction to construct (lt,k)∞ t=0 for each k ∈ N is completed. Next, let t > m and notice that q q q�lt,k − lm,k�p = �lt,k − lt−1,k�p + . . . + �lm+1,k − lm,k�p ≤ + . . . + ≤ . 2k+t 2k+m+1 2k+m Therefore (lt,k)∞ t=0 is a Cauchy sequence in V , for each k. Let lim = ∈ V . Now notice t→∞ lt,k lk that (1) Since for every t, we have lt,k(sk) = fk(sk) ≠ 0, then lk(sk) = t→∞ lt,k ) = fk(sk) ≠ 0 lim (sk (2) Since for t > j and j ≠ k, we have lt,k(sj ) = 0, then lk(sj ) = lim (sj ) = 0. t→∞ lt,k (3) Since �lt,k − l0,k�p ≤ � then �lk − fk�p = lim �lt,k − l0,k�p ≤ q . 2k t→∞ 2k Thus, (fk)k∈N is a normalized basic sequence with basis constant K ≤ 8−2� such that [f1, f2, . . .] is complemented in �p with a projection P : �p with norm ��P �� 4 ≤ −9� 8−2� and→ �p 4−33� ∞ ∞ q δ =∑ �lk − fk�p ≤∑ = q. 2k k=1 k=1 Finally 8Kδ��P �� ≤ 8q (8−2� ) ( 8−2� ) < 512q < 1. By the principle of small pertubation 4−9� 4−33� [16, Theorem 4.5] the sequence (lk)k∈N is equivalent to (fk)k∈N and [l1, l2, . . .] is complemented in �p. Finally define lsk = lk for k ∈ N. 58 Remark 5.5. Since (fk)k∈N is equivalent to (lsk )k∈N then the convergence of ∑∞ k=1 aklsk im­ plies the convergence of ∑∞ k=1 a2kls2k and the convergence of ∑ ∞ k=1 a2k−1ls2k−1 , by remark 5.3. Therefore ∑∞ k=1 aklsk = k=1 a2kls2k k=1 a2k−1ls2k−1 .∑∞ +∑∞ Proposition 5.6. Let V be an infinite dimensional closed subspace of �p, p ∈ [1, ∞[. There exists 0 ≠ h ∈ V ∖Z(V ). Proof. Consider any lk from Lemma 5.4. Notice that any lk ∈ V ∖Z(V ). Corollary 5.7. There is no infinite dimensional closed subspace V of �p, p ∈ [1, ∞[, such that V ⊂ Z(�p) ∪ {0}. Corollary 5.8. Let V be an infinite dimensional closed subspace of �p. Then V ∖ Z(V ) is dense in V . fProof. Let 0 ≠ f ∈ V . Define f1 = lf l . We can start the proof of Lemma 5.2 using this f1. p Consider the proof of Lemma 5.4. For a sufficiently small q (independent of �f �p), we found a l1 ∈ V ∖Z(V ) such that �f1 − l1�p < 2 � 1 then �f � �f − �f �p l1� < pq . 2 Now 0 ≠ �f �pl1 ∈ V ∖Z(V ). 5.2 The case: X = c0 or �∞ This section shall provide the definitive answer to Question 5.1 by showing that �∞ does not contain infinite dimensional Banach subspaces every nonzero element of which has only a finite number of zero coordinates. In order to achieve this we shall need to obtain a sequence (lsk )k∈N similar to that from Lemma 5.4 (see Lemma 5.14). Despite losing the hypothesis of the closed span of (lsk )k∈N being complemented, we gain the property lsk (sk) = 1, obtaining still a basic sequence. Definition 5.9. Let V be an infinite dimensional closed subspace of �∞. Let s ∈ N and define �f �∞ Vs = {f ∈ V, f ≠ 0, �f(s)� ≥ } . 2 Lemma 5.10. Let V be an infinite dimensional closed subspace of �∞. For every K ⊂ V , K ≠ {0}, there exists s ∈ N such that Vs ∩K ≠ ∅. 59 Proof. Let f ∈ K, f ≠ 0. Since �f �∞ = supk∈N �f(k)� there is s ∈ N such that �f(s)� ≥ lf2 l∞ . So f ∈ Vs ∩K. Lemma 5.11. Let V be an infinite dimensional closed subspace of �∞. There exist an in­ creasing sequence of natural numbers (nk)k∈N and a basic sequence (fnk )k∈N ⊂ V with: (1) fnk (nk) = 1, (2) fnj (ni) = 0 for j > i, and (3) �fnk �∞ ≤ 2 for every k ∈ N. Proof. This proof is a variation of Mazur’s lemma ([16, Proposition 4.1]). ∞ Let q1 = 1 and qi ∈ ]0, 1[ be such that ∏(1 + qi) <∞. i=1 By Lemma 5.10, there exists s ∈ N such that Vs = Vs ∩ V ≠ ∅. Let n1 = min{s ∈ N, Vs ≠ ∅} and let f1 ∈ Vn1 . Define f1 fn1 = ) . f1(n1 Notice that �f1�∞ fn1 (n1) = 1 and 1 ≤ �fn1 �∞ = �f1(n1)� ≤ 2. Consider the projection πn1 : V → C, πn1 (f) = f(n1). Let W1 = ker(πn1 ). Since codim(W1) in V is finite then dim(W1) = ∞, by Lemma 5.10 there exists s ∈ N such that Vs ∩W1 ≠ ∅. Let n2 = min{s ∈ N, Vs ∩W1 ≠ ∅}. Since Vs ⊃ Vs ∩W1 then n2 ≥ n1. Now for every f ∈ W1, f(n1) = 0 then Vn1 ∩W1 = ∅ then n2 > n1. Next, let f2 ∈ Vn2 ∩W1 and define f2 fn2 = ) . f2(n2 Notice that �f2�∞ fn2 (n2) = 1, fn2 (n1) = 0, and 1 ≤ �fn2 �∞ = �f2(n2)� ≤ 2. 60 Next, for a1, a2 ∈ C, �a1fn1 + a2fn2 �∞ ≥ �πn1 (a1fn1 + a2fn2 )� = �a1� and 1 + q1 = 2 ≥ �fn1 �∞, so �a1fn1 + a2fn2 �∞(1 + q1) ≥ �a1��fn1 �∞ = �a1fn1 �∞. Consider now the compact set S1(⟨fn1 , fn2 ⟩) and let {y1, . . . , yk} ⊂ S1(⟨fn1 , fn2 ⟩) be such that if y ∈ S1(⟨fn1 , fn2 ⟩) then there exists yi such that �y − yi�∞ < �2 2 . Consider {φ1, . . . , φk} ⊂ S1(V ∗) such that φi(yi) = 1. Take πn2 : V → C, Tn2 (f) = f(n2). Let k W2 =⋂ ker(φi) ∩ ker(πn2 ) ∩W1. i=1 Since codim(ker(φi)), codim(ker(πn2 )), and codim(W1) are finite in V then codim(W2) is finite and dim(W2) =∞. By lemma 5.10 there exists s ∈ N such that Vs ∩W2 ≠ ∅. Let n3 = min{s ∈ N, Vs ∩W2 ≠ ∅}. Since Vs ∩W1 ⊃ Vs ∩W2 then n3 ≥ n2. Now, for all f ∈ W2, f(n2) = 0 then Vn2 ∩W2 = ∅ then n3 > n2 . Next, let f3 ∈ Vn3 ∩W2 and define f3 fn3 = ) . f3(n3 Notice that �f3�∞ fn3 (n3) = 1, fn3 (n2) = fn3 (n1) = 0, and 1 ≤ �fn3 �∞ = �f3(n3)� ≤ 2. Now, let y ∈ S1(⟨fn1 , fn2 ⟩). Notice that �y + λfn3 �∞ ≥ �yi + λfn3 �∞ − �yi − y�∞ ≥ �yi + λfn3 �∞ − q2 (for some i ∈ {1, . . . , k}) 2 ≥ φi(yi + λfn3 ) − q2 2 ) − q2≥ φi(yi 2 ≥ 1 − q2 ≥ 1 . 2 1 + q2 Thus, for every y ∈ S1(⟨fn1 , fn2 ⟩) and any λ ∈ C we have �y + λfn3 �∞(1 + q2) ≥ �y�∞. 61 Then �a1fn1 + a2fn2 + a3fn3 �∞(1 + q2) ≥ �a1fn1 + a2fn2 �∞ for all a1, a2, a3 in C. We can repeat the procedure to build fn4 , fn5 , . . . satisfying �a1fn1 + . . . + akfnm �∞(1 + qm−1) . . . (1 + qk) ≥ �a1fn1 + . . . + akfnk �∞ for every a1, . . . , am ∈ C and m ≥ k and by Banach’s criterion (fnk )k∈N ⊂ V is a basic sequence. Note that (fnk )k∈N satisfies the desired conditions. Lemma 5.12. Let g1, g2 ∈ �∞ and let (mk)k∈N be an increasing sequence of natural numbers. 1There exists (mk)k∈N ⊂ (mk)k∈N such that 1(1) There exists lim (mk) = L1, k→∞ g1 1(2) There exists lim (mk) = L2, and k→∞ g2 1 1(3) m2 > m1 > m2 > m1. Proof. The sequence (g1(mk))k∈N is bounded since g1 ∈ �∞, therefore there is a subsequence 0 0(mk)k∈N ⊂ (mk)k∈N and L1 ∈ C such that lim (mk) = L1, and by same reasoning, there is a k→∞ g1 1 0 1 1subsequence (m )k∈N ⊂ (m )k∈N and L2 such that lim (mk) = L2. Therefore lim (mk) = k k k→∞ g2 k→∞ g1 1 1L1 and lim (m ) = L2. Removing, if necessary, the first two terms in the sequence (m )k∈N k→∞ g2 k k 1 1we may assume that m2 > m1 > m2 > m1. Lemma 5.13. Let V be an infinite dimensional closed subspace of �∞ and let (nk)k∈N be as in Lemma 5.11. For every (mk)k∈N ⊂ (nk)k∈N there exist (tk)k∈N ⊂ (mk)k∈N and basic sequence (htk )k∈N ⊂ V satisfying a) htk (ts) = 0 for s < k, b) htk (tk) = 1, c) �htk �∞ ≤ 8, and d) s lim →∞ htk (ts) = 0. Proof. Let (fnk )k be as in Lemma 5.11. Define g1 = fm1 − fm1 (m2)fm2 and g2 = fm2 . Notice that g1(m1) = 1, g1(m2) = 0, g2(m1) = 0 and g2(m2) = 1. Now by Lemma 5.12 there exist 1(mk)k∈N ⊂ (mk)k∈N such that 1 1 1 1lim (m ) = L1, lim (m ) = L2, and m > m > m2 > m1.k k 2 1 k→∞ g1 k→∞ g2 We now have the following possibilities. 62 (1) If L1 = 0, let h1 = g1. Notice that, since �fmi �∞ ≤ 2 (1 ≤ i ≤ 2) we have �h1�∞ ≤ 6. Notice also that h1(m1) = 1. (2) If L1 ≠ 0 and L2 = 0, let h1 = g2. We have �h1�∞ ≤ 2 and h1(m2) = 1. − L1(3) If L1 ≠ 0, L2 ≠ 0 and �L1� ≤ �L2�, define h1 = g1 g2. Notice that �h1�∞ ≤ �g1�∞ + �L1� L2 �g2�∞ ≤ 8. Also, h1(m1) = 1.�L2� − L2 (4) Finally, if L1 ≠ 0, L2 ≠ 0 and �L2� ≤ �L1�, let h1 = g2 g1, having now that �h1�∞ ≤ �L2� L1 �g2�∞ + �g1�∞ ≤ 8. Also, note that h1(m2) = 1.�L1� Next, if h1(m1) = 1, define t1 = m1 and, if h1(m1) ≠ 1, then h1(m2) = 1 and we let t1 = m2. In 1any case, note that lim h1(mk) = 0. Let us now suppose that, by induction, we have already k→∞ defined i 1(1) (mk)k∈N ⊂ ⋯ ⊂ (mk)k∈N ⊂ (mk)k∈N with i i i−1 i−1 1 1 m > m > m > m > . . . > m > m > m2 > m1,2 1 2 1 2 1 j−1 j−1(2) t1 = m1 or m2 and tj = m1 or m2 , 2 ≤ j ≤ i. (3) hj ∈ V , 1 ≤ j ≤ i, verifying conditions a), b) and c) of this lemma, and j(4) lim hj (m ) = 0, 1 ≤ j ≤ i. k→∞ k Next, repeat the construction of h1 in order to obtain hi+1 by means of fmi i, fm instead 1 2 of fm1 , fm2 . iUsing the sequence (m )k∈N, instead of (mk)k∈N in the previous construction, we obtain i+1 i k (mk )k∈N ⊂ (mk)k∈N such that i+1 i+1 i i i+1 m > m > m > m and lim hi+1(m ) = 0.2 1 2 1 k k→∞ i i i iDefine now ti+1 = m or m , depending on whether hi+1(m ) = 1 or hi+1(m ) = 1, as we 1 2 1 2 previously did for t1. Therefore we have hi+1(ti+1) = 1. Next, since hi+1 is a linear combination of fm , fm , and i i 1 2 i i i−1 i−1 1 1m > m > m > m > . . . > m > m > m2 > m1,2 1 2 1 2 1 63 � j−1 j−1we obtain that hi+1(m1) = hi+1(m2) = hi+1(m1 ) = hi+1(m2 ) = 0 (for 2 ≤ j ≤ i), but t1 = m1 j−1 j−1or m2, tj = m1 or m2 (for 2 ≤ j ≤ i), which implies that hi+1(tj ) = 0 for 1 ≤ j ≤ i. iFinally, notice that (ts)∞ ⊂ (m )k∈N, thus lim (ts) = 0 (for every i ∈ N).s=i+1 k s→∞ hi Notice that (fmk )k∈N is a basic sequence as subsequence of the basic sequence (fnk )k∈N. Notice also that hk is a linear combination of fmk−1 and fmk−1 , h1 is a linear combination of 1 2 fm1 and fm2 and mk−1 > mk−1 > . . . > m1 > m1 > m2 > m1 for every k. Therefore (hk)k∈N is2 1 2 1 a block sequence of the basic sequence (fmk )k∈N. Therefore (hk)k∈N is also a basic sequence. Finally, let htk = hk. Lemma 5.14. Let V be an infinite dimensional closed subspace of �∞ and let (nk)k∈N be as in Lemma 5.11. For every (mk)k∈N ⊂ (nk)k∈N there exist (sk)k∈N ⊂ (mk)k∈N and a basic sequence (lsk )k∈N ∈ V , satisfying a) lsk (sk) = 1, b) lsk (sj ) = 0, for j ≠ k. c) �lsk �∞ ≤ 9, for every k ∈ N. Proof. Consider (tk)k∈N ⊂ (mk)k∈N and (htk )k∈N ⊂ V as in Lemma 5.13. Let K be the basis 1constant of the basic sequence (htk )k∈N and let 0 < q < 2K . (Recall that K is always equal or bigger than 1, therefore q < 1 ). Let s1 = t1. Suppose defined, by induction, {s1, . . . , sn} ⊂ {t1, t2, . . .}. Since lim �hs1 (tj )� + . . . + �hsn (tj )� = 0, there exists sn+1 ∈ {t1, t2, . . .}, sn+1 > sn, j→∞ such that q�hs1 (sn+1)� + . . . + �hsn (sn+1)� ≤ . 2n+1 × 8 The induction to construct (sk)k∈N ⊂ (mk)k∈N is completed. Now define l0,k = hsk . Notice that l0,k(sk) = 1 and l0,k(sj ) = 0 for sj ∈ {s1, ..., sk} ∖ {sk}. Define l1,k = l0,k − l0,k(sk+1)hsk+1 . Notice that • l1,k(sk) = 1 and l1,k(sj ) = 0 for sj ∈ {s1, ..., sk+1} ∖ {sk}, • since �hsk (sk+1)� ≤ 2k+1×8 then �l1,k(sj )� ≤ �l0,k(sj )� + �hsk+1 (sj )� = �hsk (sj )� + �hsk+1 (sj)� for every j ∈ N. 64 � � � � � • �l1,k − l0,k�∞ = �l0,k(sk+1)��hsk+1 �∞ ≤ 2k+1×8 × 8 = 2k+1 Suppose we have already defined, by induction, l0,k, l1,k, . . . , lt,k ∈ V such that • ln,k(sk) = 1 for 0 ≤ n ≤ t, • ln,k(sj) = 0 for sj ∈ {s1, ..., sk+n} ∖ {sk} and 0 ≤ n ≤ t, • �ln,k(sj )� ≤ �hsk (sj )� + �hsk+1 (sj )� + . . . + �hsk+n (sj )�, for every j ∈ N and 0 ≤ n ≤ t, • �ln,k − ln−1,k�∞ ≤ 2k+n for 1 ≤ n ≤ t, Next, define lt+1,k = lt,k − lt,k(sk+t+1)hsk+t+1 . Notice that • lt+1,k(sk) = 1, • lt+1,k(sj ) = 0 for sj ∈ {s1, ..., sk+t+1} ∖ {sk}, • since �lt,k(sk+t+1)� ≤ �hsk (sk+t+1)� + �hsk+1 (sk+t+1)� + . . . + �hsk+t (sk+t+1)� ≤ q �hs1 (sk+t+1)� + �hs2 (sk+t+1)� + . . . + �hsk+t (sk+t+1)� ≤ 2k+t+1 × 8 then �lt+1,k(sj )� ≤ �lt,k(sj )� + �hsk+t+1 (sj )� for every j ∈ N and by induction hypothesis �lt+1,k(sj )� ≤ �hsk (sj )� + �hsk+1 (sj )� + . . . + �hsk+t+1 (sj )�, for every j ∈ N. • �lt+1,k − lt,k�∞ = �lt,k(sk+t+1)��hsk+t+1 �∞ ≤ 2k+t+1×8 × 8 = 2k+t+1 )∞The induction to construct (lt,k t=0 ⊂ V , for every k ∈ N, is completed. Now q q�l0,k�∞ + �l1,k − l0,k�∞ + �l2,k − l1,k�∞ + . . . ≤ �l0,k�∞ + + + . . . ≤ �l0,k�∞ + q. 2k+1 2k+2 Thus, for each k ∈ N, the series lim = ) + . . . is absolutely t→∞ lt,k l0,k + (l1,k − l0,k) + (l2,k − l1,k and coordinatewise convergent to some lk ∈ V . Notice that lt,k(sk) = 1 for every t then t lim →∞ lt,k(sk) = lk ) = 1. Next lt,k(sj ) = 0 for t > j and j ≠ k then lim (sj ) = lk ) =(sk t→∞ lt,k (sj 0. Now, lt,k − l0,k = (lt,k − lt−1,k) + . . . + (l1,k − l0,k) then q q q ≤ q�lt,k − l0,k�∞ ≤ + + . . . + , 2k+t 2k+t−1 2k+1 2k 65 � � . then lim �lt,k − l0,k�∞ = �lk − hsk �∞ ≤ q , for every k ∈ N, so t→∞ 2k �lk�∞ ≤ �hsk �∞ + 2 q k ≤ 8 + 1 = 9. Since hsk (sk) = 1 then �hsk �∞ ≥ 1 and we have lk − hsk ≤ q �hsk �∞ �hsk �∞ 2k∞ ∞ lk − hsk ∞ q Then δ :=∑ � � ≤∑ = q. k=1 �hsk �∞ �hsk �∞ ∞ k=1 2 k Now the normalized sequence ( lh h sk sk l∞) as a block basis of the basic sequence (htk )k∈N k∈N is also a basic sequence with basic constant K ′ ≤ K. Then 2K ′ δ ≤ 2Kδ ≤ 2Kq < 1. lkBy the principle of small pertubation [16, Theorem 4.5] the sequence ( l l∞) k∈N is a basic hsk sequence equivalent to the normalized basic sequence ( lh h s s k k l∞) . Notice that (lk)k∈N is a k∈N lkblock basis of ( lhsk l∞ ) , therefore it is also a basic sequence. Finally define lsk = lk. k∈N From the previous lemma, we can now infer the following. Proposition 5.15. Let V be an infinite dimensional closed subspace of �∞. There exists 0 ≠ h ∈ V ∖Z(V ). Proof. Consider lsk from Lemma 5.14. We have that lsk ∈ V ∖Z(V ). Corollary 5.16. There is no infinite dimensional closed subspace V of �∞, such that V ⊂ Z(�∞) ∪ {0}. As a consequence of Lemma 5.14 we also have the following result, whose proof is simple. Corollary 5.17. Let V be an infinite dimensional closed subspace of c0. Then V ∖Z(V ) is dense in V . Proof. Let (nk)k be as in Lemma 5.11. Every 0 ≠ f ∈ V ⊂ c0 satisfies lim f(nk) = 0. Let q > 0. k→∞ There exists (mk)k∈N ⊂ (nk)k∈N such that (f(mk))k∈N ∈ l1 and �(f(mk))k∈N�1 ≤ q . 9 By Lemma 5.14, there exist (sk)k∈N ⊂ (mk)k∈N and lsk ∈ V such that 66 a) lsk (sk) = 1, b) lsk (sj ) = 0, for j ≠ k. c) �lsk �∞ ≤ 9 for every k ∈ N. Notice that �f(s1)ls1 �∞ + �f(s2)ls2 �∞ + . . . ≤ (�f(s1)� + �f(s2)� + . . .) 9 ≤ q. Therefore f − f(s1)ls1 − f(s2)ls2 − . . . converges absolutely and coordinatewise to some g ∈ V . Notice that for every k ∈ N g(sk) = f(sk) − f(s1)ls1 (sk) − f(s2)ls2 (sk) − . . . = f(sk) − f(sk)lsk (sk) = 0 and �g − f �∞ ≤ q. Theorem 5.18. Let V be an infinite dimensional closed subspace of �p, p ∈ [1, ∞]. There exists an infinite dimensional closed subspace of �p, p ∈ [1, ∞] inside V ∖Z(V ) ∪ {0}. Proof. By Lemmas 5.4 and 5.14, there is an increasing sequence of natural numbers (sk)k∈N and a sequence (lsk )k∈N ⊂ V such that a) lsk (sk) ≠ 0, b) lsk (sj ) = 0, for j ≠ k. Let W = ⟨ls2 , ls4 , ls6 , . . .⟩ and notice that every f ∈ W satisfies f(s2k−1) = 0 for every k ∈ N. Since convergence in norm implies coordinatewise convergence in �p, p ∈ [1, ∞] then for every f ∈ W , we obtain f(s2k−1) = 0 for every k ∈ N. Notice that {l2k ∈ W, k ∈ N} is a linear independent set then W is a infinite dimensional closed subspace of V with W ⊂ V ∖Z(V ) ∪ {0}. Remark 5.19. If V of theorem 5.18 is also an algebra with the coordinatewise product then every element of the closed subalgebra generated by W has zeros at coordinates s1, s3, s5 . . . Corollary 5.20. Let V be an infinite dimensional closed subspace of �p, p ∈ [1, ∞[. Then the infinite dimensional closed subspace W ⊂ V ∖ Z(V ) ∪ {0}, obtained in Theorem 5.18, is complemented in �p. Proof. Notice that the sequence (lsk )k∈N ⊂ V used in the proof of Theorem 5.18 is the basic sequence constructed in lemma 5.4, when p ∈ [1, ∞[. Thus, [ls1 , ls2 , . . .] is complemented in �p. Since W = [ls2 , ls4 , ls6 , . . .] is complemented in [ls1 , ls2 , . . .] by [ls1 , ls3 , ls5 , . . .], by remark 5.5. We got the result. 67 List of Symbols and Notation Chapters 2 and 3: Ck – The set of column vectors with k complex entries. Mk – The set of complex matrices of order k. Pk – The set of positive semidefinite Hermitian matrices of order k. X ⊗ Y – The Kronecker product of the matrices X, Y . Ck ⊗Cm – The tensor product space of Ck and Cm. Mk ⊗Mm – The tensor product space of Mk and Mm. Mk1 ⊗ . . . ⊗Mkn – The tensor product space Mk1 ⊗ (Mk2 ⊗ . . . ⊗Mkn ). Id – The Identity matrix. VMkW – The set {V XW,X ∈ Mk}, where V, W ∈ Mk are orthogonal projections. tr(X) – The trace of the matrix X. X t – The transpose of the matrix X. X – The matrix whose entries are the complex conjugate of the entries of the matrix X. X ∗ – The conjugate transpose of the matrix X, i.e., X ∗ = X t . ⟨X, Y ⟩ – The trace inner product of the square matrices X, Y , i.e., tr(XY ∗). I(γ) – The image (or the range) of the matrix γ. T ∗ :WMmW → VMkV – The adjoint of T : VMkV →WMmW with respect to ⟨X, Y ⟩. �X�2 – The spectral norm of the matrix X ∈ Mk. X+ – The pseudo-inverse of the matrix X. R ′ +R – The sum of the spaces R ′ ,R. R ′ ⊕R – The direct sum of the spaces R ′ ,R. R ′ ⊥ R – The orthogonality of the subspaces R ′ ⊂ Mk and R ⊂ Mk with respect to ⟨X, Y ⟩. L�R – The restriction of the map L : VMkV → VMkV to R ⊂ VMkV . FA :Mm →Mk – The map FA(X) = ∑i n =1 tr(BiX)Ai, where A = ∑n i=1 Ai ⊗Bi ∈ Mk ⊗Mm. GA :Mk →Mm – The map GA(X) = ∑n i=1 tr(AiX)Bi, where A = ∑n i=1 Ai ⊗Bi ∈ Mk ⊗Mm. tx – The transposition of the column vector x ∈ Ck. x – The column vector whose entries are the complex conjugate of the entries of x. ⟨x, y⟩ – The usual inner product of the column vectors x, y ∈ Ck, i.e., ⟨x, y⟩ = xty. 69 At2 – The partial transposition of A = ∑n i=1 Ai ⊗Bi ∈ Mk ⊗Mm, i.e., At2 = ∑n i=1 Ai ⊗Bi t . det(γ) – The determinant of the matrix γ. Chapter 4 and 5: C (D)– The set of real-valued continuous functions on a topological space D. C�(D) – The subset of C (D) formed by the functions that attain the maximum exactly once in D. dim(V ) – The dimension of the vector space V . Sk – The Euclidean sphere of radius 1 of Rk+1. ∂(A) – The frontier of A ⊂ S2. int(A) – The interior of A ⊂ S2. y(j) – The j−th coordinate of the sequence y, i.e., y = (y(j))j∈N . (mk)k∈N ⊂ (nk)k∈N – This symbol means that (mk)k∈N is a subsequence of (nk)k∈N. ⟨v1, v2, . . .⟩ – The linear space spanned by (vk)k∈N. [v1, v2, . . . ] – The closed linear space spanned by (vk)k∈N. S1(W ) – The sphere of radius 1 of the normed set W , i.e., S1(W ) = {w ∈ W, �w� = 1}. 70 Bibliography [1] R. M. Aron, L. Bernal-González, D. Pellegrino, and J. B. Seoane-Sepúlveda, Lineability: The search for linearity in Mathematics, Monographs and Research Notes in Mathematics, Chapman & Hall/CRC, Boca Raton, FL, 2015, ISBN 978-1-48-229909-0. [2] R. M. Aron, V. I. Gurariy, and J. B. Seoane-Sepúlveda, Lineability and spaceability of sets of functions on R, Proc. Amer. Math. Soc. 133 (2005), no. 3, 795–803. [3] R. Augusiak, M. Demianowicz, and P. 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Fields, Optimal state-determination by mutually unbiased measurement, Annals of Physics 191 (1989), 363-381. 73 Tesis Daniel Cariello Agradecimientos Contents Resumen Abstract Introducción 1. Introduction 2. Some Results from the Perron-Frobenius Theory Two theorems from the Perron-Frobenius Theory Two related properties: completely reducibility and the decomposition property 3. Completely Reducible Maps in Quantum Information Theory Main Theorems: The Complete Reducibility of FAGA: MkMk Main Theorem for PPT matrices Main Theorems for SPC matrices and Matrices Invariant under Realignment Applications to Quantum Information Theory The Separability Problem Extension of Mutually Unbiased Bases Remarks Some Remarks on our Main Theorems A Remark on the Application to the Separability Problem 4. An application of Borsuk-Ulam Theorem Main Result An Infinite Dimensional Example Open Problem 5. Basic Sequences in p spaces The case: X =p, p[1,[ The case: X = c0 or