MAJORIZATION AND THE SCHUR-HORN THEOREM

Transcription

1 MAJORIZATION AND THE SCHUR-HORN THEOREM A Thesis Submitted to the Faculty of Graduate Studies and Research In Partial Fulfillment of the Requirements for the Degree of Master of Science In Mathematics University of Regina By Maram Albayyadhi Regina, Saskatchewan January 2013 c Copyright 2013: Maram Albayyadhi

2 UNIVERSITY OF REGINA FACULTY OF GRADUATE STUDIES AND RESEARCH SUPERVISORY AND EXAMINING COMMITTEE Maram Albayyadhi, candidate for the degree of Master of Science in Mathematics, has presented a thesis titled, Majorization and the Schur-Horn Theorem, in an oral examination held on December 18, The following committee members have found the thesis acceptable in form and content, and that the candidate demonstrated satisfactory knowledge of the subject material. External Examiner: Supervisor: Committee Member: Dr. Daryl Hepting, Department of Computer Science Dr. Martin Argerami, Department of Mathematics and Statistics Dr. Douglas Farenick, Department of Mathematics and Statistics Chair of Defense: Dr. Sandra Zilles, Department of Computer Science

3 Abstract We study majorization in R n and some of its properties. The concept of majorization plays an important role in matrix analysis by producing several useful relationships. We find out that there is a strong relationship between majorization and doubly stochastic matrices; this relation has been perfectly described in Birkhoff s Theorem. On the other hand, majorization characterizes the connection between the eigenvalues and the diagonal elements of self adjoint matrices. This relation is summarized in the Schur-Horn Theorem. Using this theorem, we prove versions of Kadison s Carpenter s Theorem. We discuss A. Neumann s extension of the concept of majorization to infinite dimension to that provides a Schur-Horn Theorem in this context. Finally, we detail the work of W. Arveson and R.V. Kadison in proving a strict Schur-Horn Theorem for positive trace-class operators. i

4 Acknowledgments Throughout my studying, I could have not done my project without the support of my professors whom I insist on thanking even though my words cannot adequately express my gratitude. Dr. Martin Argerami, I would like to thank you from the bottom of my heart for all the support and the guidance that you provided for me. Dr. Shaun Fallat and Dr. Remus Floricel, if it were not for your classes, I would not have learned as much about my field. I m also honored to thank all my amazing colleagues in the math department, especially my friend Angshuman Bhattacharya. To my father Ibrahim Albayyadhi and my Mother Suad Bakkari, words cannot express my love for you. Your prayers, belief in me and encouragement are the main reasons for my success. If I kept on thanking you all of my life, I could not pay you back. To my husband, Dr. Hadi Mufti, you always make it easier for me whenever I face obstacles; you have always been the wind beneath my wings. Finally, I would like to thank the one who kept wiping my tears on the hard days and saying, Mom... don t give up my son Yazan. ii

5 Contents Abstract i Acknowledgments ii Table of Contents iii 1 Preliminaries Majorization Doubly Stochastic Matrices Doubly Stochastic Matrices and Majorization The Schur-Horn Theorem in the Finite Dimensional Case The Pythagorean Theorem in Finite Dimension The Schur-Horn Theorem in the Finite Dimensional Case A Pythagorean Theorem for Finite Doubly Stochastic Matrices The Carpenter Theorem in the Infinite Dimensional Case The Subspaces K and K both have Infinite Dimension One of the Subspaces K, K has Finite Dimension A Pythagorean Theorem for Infinite Doubly Stochastic Matrices.. 47 iii

6 4 A Schur-Horn Theorem in Infinite Dimensional Case Majorization in Infinite Dimension Neumann s Schur-Horn Theorem A Strict Schur-Horn Theorem for Positive Trace-Class Operators Conclusion and Future Work 66 iv

7 Chapter 1 Preliminaries In this chapter we provide some basic information about majorization and some of it properties, which we will use later. The material in this chapter is basic and can be found in many matrix analysis books [4, 7]. 1.1 Majorization Let x = (x 1,, x n ) R n. Let x = (x 1,, x n) and x = (x 1,, x n) denote the vector x with it s coordinates rearranged in increasing and decreasing orders respectively. Then x i = x n i+1, 1 i n. Definition 1.1. Given x, y in R n, we say x is majorized by y, denoted by x y, 1

8 if and x i = x i y i, 1 k n (1.1) y i. (1.2) Example 1.2. If x i [0, 1], and n x i = 1, then we have ( 1 n,. 1 n ) (x 1,, x n ) (1, 0,, 0). Proposition 1.3. If x y and y x, then there exists a permutation matrix P such that y = Px. Proof. Assume first that x, y are rearranged in decreasing order,i.e. x 1 x 2 x n, y 1 y 2 y n. We proceed by induction on k. For k = 1 we get x 1 y 1, and y 1 x 1 from the majorization equation (1.1), which implies x 1 = y 1. By induction hypothesis we assume that x i = y i, for i = 1,, k and we will show that it works up to k + 1. From ( 1.1) x 1 + x x k + x k+1 y 1 + y y k + y k+1, and by our assumption x 1 + x x k = y 1 + y y k, so if we cancel them we get x k+1 y k+1, as the roles of x and y can be reversed y k+1 x k+1, and this implies x k+1 = y k+1. In general, if we write x and y for the non-increasing rearrangements, there exist permutations P 1 and P 2 such that x = P 1 x, y = P 2 y. By the first part of the proof, x = y, i.e. x = P 1 1 P 2 y. Although majorization is defined through non-increasing rearrangements, it can 2

9 also be done by non-decreasing ones: Proposition 1.4. If x, y R n, then x y if and only if and x i = x i y i, 1 k n (1.3) y i (1.4) Proof. For equation(1.4), we have n x i = n y i, since x y, but n x i = n x i, so n x i = n x i = n y i = n y i, and n x i = n y i. And for equation (1.3), we know that x i = x n i+1, 1 i n. So Then x i = x n i+1. x i = n k x l = x l l=n k+1 l=1 l=1 n k = tr (x) l=1 n k tr (y) l=1 x l x l y l n k = y l y l = l=1 l=1 l=n k+1 y l = y i. 3

10 1.2 Doubly Stochastic Matrices There is a deep relation between majorization and doubly stochastic matrices; we will discuss that relation in this section. Definition 1.5. Let B = (b ij ) be a square matrix. We say B is doubly stochastic if : b ij 0 i, j, n b ij = 1 j, n j=1 b ij = 1 i. Proposition 1.6. The set of square doubly stochastic matrices is a convex set and it is closed under multiplication and the adjoint operation. But it is not a group. Proof. If t 1,, t r [0, 1] with r j=1 t j = 1 and P 1,, P r are permutations, let A = r j=1 t jp j. Clearly every entry of A is non-negative. Also, A kl = k=1 = = r t j (P j ) kl k=1 j=1 r j=1 t j (P j ) kl k=1 r t j = 1, l. j=1 A similar computation shows that n l=1 A kl = 1 for all k. If A, B are doubly stochastic matrices then AB is also doubly stochastic. Indeed, 4

11 the sum over the rows is (AB) ij = j=1 = = A ik B kj j=1 k=1 k=1 A ik j=1 A ik = 1. k=1 B kj And the same thing can be done for the columns, which shows that AB is also doubly stochastic. Also it is clear that if A doubly stochastic, then so is its adjoint A. The class of doubly stochastic matrices is not a group since not every doubly stochastic matrix is invertible; for example if we take the 2 2 matrix, with all its entries equal to 1, this matrix is doubly stochastic and its determinant is zero. 2 Proposition 1.7. Every permutation matrix is doubly stochastic and is an extreme point of the convex set of all doubly stochastic matrices. Proof. A permutation matrix has exactly one entry +1 in each row and in each column and all other entries are zero. So it is doubly stochastic. Now let A = α 1 B + α 2 C, with A a permutation matrix such that α 1, α 2 (0, 1), α 1 + α 2 = 1, and B, C are doubly stochastic matrices. Then every entry of B, C that corresponds to the zero element a ij = 0 of A must be zero; indeed, 0 = a ij = α 1 b ij + α 2 c ij. As α 1, α 2 both are non zero, and B, C are both nonnegative, we have b ij = c ij = 0. Hence, nonzero entries must be all +1, A = B = C. This shows that every permutation matrix is an extreme point of the set of doubly stochastic matrices. 5

12 1.3 Doubly Stochastic Matrices and Majorization Theorem 1.8. A matrix A M n (R) is doubly stochastic if and only if Ax x, for all x R n. Proof. For the implication ( ), assume Ax x for all vectors x. Then Ax = x. From the definition of majorization. If we choose x to be e j, where e j is the vector e j = (0,, 0, 1, 0,, 0), 1 j n, then a 11 a 1n 0 a 1j =. 1. a n1 a nn 0 a nj 0 Then min{a 1j,, a nj } min{1, 0} = 0, so a kj 0 for all k. Also this implies, as j was arbitrary, that the sum over each column is 1. To show the sum over the rows is also 1 we use the vector e = n j=1 e j; we get a 11 a 1n 1 n j=1... = a 1,j 1... a n1 a nn 1 n j=1 a n,j 1 Then n j=1 a ij = 1 for all i, since max{ n j=1 a ij : j} 1, min{ n j=1 a ij; j} 1. For the other direction ( ), let A be doubly stochastic, and let y = Ax. To prove 6

13 y x, we first show that we can assume x and y have their entries in non-increasing order; this because x = Px, and y = Qy for some permutation matrices P and Q, so Qy = APx y = Q 1 APx, y = Bx. where Q 1 AP = B is doubly stochastic, since the permutation matrices are doubly stochastic, and the product of doubly stochastic matrices is doubly stochastic by Proposition 1.6. For any k {1,, n} we have y j = j=1 b ji x i. (1.5) j=1 Let s i = k j=1 b ji, then 0 s i 1, n s i = k and y j = j=1 s i x i. Then y j x j = j=1 j=1 s i x i x i. (1.6) 7

14 By adding and subtracting n s ix k from equation (1.6), y j j=1 x j = j=1 = = = s i x i s i x i s i x i + s i x i + = = x i + ( s i x i + (k i=k+1 i=k+1 (1 s i )x i + s i x i s i x i s i )x k s i )x k, since k = x i + x i + x k x k (x i x k )s i + i=k+1 (s i 1)(x i x k ) + 0. s i x k s i s i x k (1 s i )x k (x i x k )s i i=k+1 i=k+1 s i x k So k j=1 y j k j=1 x j for all k. When k = n, y j = j=1 b ji x j j=1 = ( i b i1 )x ( i b in )x n = x j. j=1 Definition 1.9. Let A R n be a linear map. We say that A is a T-transform if there exists a [0, 1] and j, k such that Ay = (y 1,, y j 1, ay j + (1 a)y k, y j+1,, (1 a)y j + ay k, y k+1,, y n ), 8

15 i.e. A = ai + (1 a)p, where P is the transposition (jk). Theorem Given x, y R n, the following statements are equivalent: 1. x y. 2. x = T y, where T is a product of T -transforms. 3. x conv Sy = conv{py : P is a permutation}. 4. x = Ay, where A is doubly stochastic matrix. Proof. (1) = (2): We want to show that if x y, then x = (T r T 1 )y for some T-transforms T 1,, T r. We will show this is true for any n by induction. Let x, y R n. We can assume that x, y have their coordinates in decreasing order by permuting them, and each of these permutations is a product of T -transforms (because transpositions are T - transforms, and permutations are products of transpositions). So when x y, we have y n x 1 y 1. If we take k n such that y k x 1 y k 1, then x 1 = ty 1 +(1 t)y k for some t [0, 1], and we define T 1 y = (ty 1 + (1 t)y k, y 2,, y k 1, (1 t)y 1 + ty k, y k+1,, y n ). Let x = (x 2,, x n ), y = (y 2,, y k 1, (1 t)y 1 + ty k, y k+1,, y n ) R n 1. Note that the first coordinate of T 1 y is x 1. If we take off x 1 from x and T 1 y, then x = x and T 1 y = y. We will show that x y. For m such that 2 m k 1, m y j j=2 m x 1 j=2 m x j. j=2 9

16 And for k m n we have m 1 j=1 y j = = = = k 1 y j + [(1 t)y 1 + ty k ] + j=2 k 1 y j + y 1 ty 1 + ty k + j=2 m y j ty 1 (1 t)y k, j=1 m y j x 1 j=1 m x j x 1 = j=1 m 1 j=1 x j m j=k+1 m j=k+1 y j y j by adding and subtracting y k The last inequality is an equality when m = n since x y and so x y. By induction hypothesis there exist a finite number of T-transforms T 2,, T r on R n 1 such that x = (T r,, T 2 )y. We can regard each of them as T-transform on R n if we don t touch the first coordinate of any vector. Then we will have (T r T 1 )y = (T r T 2 )(x 1, y ) = (x, x ) = x. (2)= (3): It is clear that each T -transform is a convex combination of permutations. Now we have to show that a product of two convex combinations of permutations is a convex combination of permutations. For P j Q j where each of them is a permutation; t j s j 0, j, k. We have ( l ) ( m ) t j P j s k Q k = j=1 k=1 l m t j s k P j Q k. j=1 k=1 10

17 Where l m t j s k = j=1 k=1 l m t j j=1 k=1 s k = l t j = 1. j=1 (3) = (4): Is trivial, since we have x conv(py), and from Proposition 1.6 we know that a convex combination of permutation matrices is doubly stochastic. (4)= (1): This is Theorem 1.8. Definition If B is a square matrix, and P is a permutation, then we call the set { b 1P(1), b 2P(2),, b np(n) } a diagonal of B. Each element in the diagonal of B has one entry from each column and each row. Theorem (The König- Frobenius Theorem) Given a square matrix B, then every diagonal of B contains a zero element if and only if B has an i j submatrix with all entries zero for some i, j such that i + j > n. Proof. This is equivalent to Hall s Theorem. Theorem (Birkhoff s Theorem) The set of n n doubly stochastic matrices is a convex set whose extreme points are the permutation matrices. Proof. To prove this we have to show two things. First, that a convex combination of doubly stochastic matrices is doubly stochastic; this was proven in Proposition 1.7. Second we have to show that every extreme point is a permutation matrix, and for this we will show that each doubly stochastic matrix is a convex combination of a permutation matrix. This can be proved by induction on the number of nonnegative 11

18 entries of the matrix. When A has n positive entries, if A is doubly stochastic, then A is a permutation matrix. Let A be doubly stochastic. Then A has at least one diagonal with no zero entry; indeed, let [0 k l ] be a submatrix of zeros that A might have. In such case we can find permutation matrices Q 1, Q 2 such that Q 1 AQ 2 = 0 C B ; 0 is a k l submatrix with all entries zero. D Q 1 AQ 2 is doubly stochastic which means the sum of the rows of B is 1 and the sum of the columns of C is 1. i.e. n l b hi = 1, h = 1,, k, and n k c ih = 1, h = 1,, l. Also, looking at the rows and columns that intersect D, l n l c hi + d hi = 1, n k b hi + d hi = 1. h=1 h=1 Let k = n l b hi, and l = l n k c ih. 12

19 Then k = n l b ji = j=1 = n l j=1 b ji ( ) n l n k 1 d ji j=1 = n l d, where d is a positive number. Hence, k + l n. So by The König- Frobenius Theorem 1.12 at least one diagonal of A must have all its entries positive. Now suppose A is a doubly stochastic matrix with n + k non-zero entries. By the previous paragraph A has a never zero diagonal, given by some permutation Q. Let a be the minimum element of this diagonal. Clearly a < 1, because otherwise A would have a 1 in each row and column, making it a permutation, with only n non-zero entries. Let B = A aq 1 a. Then B is doubly stochastic, and the entry in B corresponding to the location of a is zero, so B has at most n + k 1 non-zero entries. By induction hypothesis B is a convex combination of permutation matrices. Since A = (1 a)b + aq, it is clear that A is a convex combination of a permutation matrices too. 13

20 Chapter 2 The Schur-Horn Theorem in the Finite Dimensional Case In this chapter we study some variants of the Pythagorean Theorem. The Pythagorean Theorem plays an important role in describing the relation between the three sides of the right triangle in Euclidean geometry. Among the variations of the Pythagorean Theorem that we will consider, some are trivial while others are not. We will find that these can be solved by using the Schur-Horn Theorem. 2.1 The Pythagorean Theorem in Finite Dimension In the following we will represent the Pythagorean Theorem (PT) in different dimensions, beginning with the classical variant. Also we will formulate the converse of the Pythagorean Theorem, which we call the Carpenter Theorem (CT)[8]. Theorem 2.1. (PT-1) If we have right triangle, such that its two sides are x,y and the angle between them is θ = π 2, then x2 + y 2 = z 2. 14

21 Although less known, the converse of (PT-1) holds. We call this the Carpenter Theorem (CT). Theorem 2.2. (CT-1) If we have a triangle with sides x, y, z, such that x 2 + y 2 = z 2, then θ = π, i.e. we 2 have a right triangle. Let {e 1, e 2 } be an orthonormal basis for R 2. Then for x R 2, we can write x as linear combination of {e 1, e 2 }, and in this case we can re-write Theorem 2.1 as Theorem 2.3. (PT-2) If x = t 1 e 1 + t 2 e 2, and x = 1, then t t 2 2 = 1. Proof. Since the norm of x is one we have 1 = x 2 = t 1 e t 2 e 2 2 = t 1 2 e t 2 2 e 2 2 = t t 2 2. Note that (PT-2) is Parseval s equality, which says that if {e j : j J} is an orthonormal basis in H, then for every x H the following equality holds: x 2 = j J x, e j 2. In what follows, we denote by P K x the orthogonal projection of x onto the subspace K. Theorem 2.4. (CT-2): If t 1, t 2 R +, and t t 2 2 = 1, then there exists x R 2 such that x = 1 and P Re1 x = t 1, P Re2 x = t 2. 15

22 Proof. Let x = t 1 e 1 + t 2 e 2. Then x 2 = t 1 e 1 + t 2 e 2 2 = t 1 e t 2 e 2 2 = t t 2 2 = 1. As P Re1 x = x, e 1 e 1, P Re1 x 2 = x, e 1 2, where x = 2 = t i e i, e 1 2 = t t i e i And the same thing for P Re2 x 2 = t 2 2. Since x = e 1 = 1, then P Re1 x = x, e 1 2 = e 1, x 2 = P Rx e 1. From this point of view, we can rephrase the (PT-2) and (CT-2) as, Theorem 2.5. (PT-3): If K is a one-dimensional subspace of R 2, then P K e P K e 2 2 = 1. Theorem 2.6. (CT-3): If t 1, t 2 R +, and t 1 + t 2 = 1, then there exists one-dimensional K R 2, such that P K e 1 2 = t 1, P K e 2 2 = t 2. Next we see that the same results hold in R n : Theorem 2.7. (PT-4): 16

23 If K is a one-dimensional subspace of R n, and {e j } n j=1 an orthonormal basis, then n j=1 P Ke j 2 = 1. Proof. We choose x = t j e j to be a unit vector for K R n, then it spans K and P K e i 2 = e i, x x 2 = e i, x 2 x 2 = e i, x 2. Then j P Ke j 2 = j e j, x 2 = x 2 by Parseval s equality. Theorem 2.8. (CT-4): If t 1,, t n [0, 1], and n j=1 t j = 1, then there exists a one-dimensional subspace K R n such that P K e j 2 = t j, j = 1,, n. Proof. Let x = n j=1 t 1 2 j e j and put K = Cx. Then P K e i = e i, x x = e i, t 1 2 j e j x = j=1 t 1 2 j e i, e j x j=1 = t 1 2 i x. So P K e i 2 = t 1 2 i x 2 = t i. In the following we are going to generalize the Pythagorean Theorem in R n, by allowing K to have different dimensions. 17

24 Theorem 2.9. (PT-5): If K is an m-dimensional subspace of R n, then n P Ke i 2 = m. Proof. If we choose f 1,, f m to be an orthonormal basis for K R n, then the projection of e i onto K is So P K e i = m e i, f j f j. j=1 P K e i 2 = = = = m e i, f j f j 2 m j=1 j=1 m e i, f j 2 j=1 e i, f j 2 m f j 2 = j=1 j=1 m 1 = m. The converse of (PT-5) would be Theorem (CT-5): If {t i } n [0, 1], and n t i = m, then there exists an m-dimensional subspace K of R n, such that P K e i 2 = t i, i = 1,, n. Suddenly, its is not so obvious how to construct K. So first we will attempt to reformulate the theorem. If K R n, P K is the orthogonal projection of R n on K, and e 1,, e n is an 18

25 orthonormal basis with (t ij ) the matrix of P K, then P K e j 2 = P K e j, P K e j = P K e j, e j = t jj, since P K = PK 2 = P K. Then n P Ke i 2 = n t i. Which we can write as n P Ke i 2 = tr (P K ). With this in mind, we can rewrite (PT,CT-5) as: Theorem (PT-6): If K is an m-dimensional subspace of R n, then tr (P K ) = m. Theorem (CT-6): If t 1,, t n [0, 1], and n j=1 t j = m, then there is K R n, such that the diagonal of P K is (t 1,..., t n ). This formulation of (CT-6) makes it clear that its proof is not going to be trivial as the previous results of (PT-CT). If we have these numbers t 1,, t n [0, 1], such that their sum is m N and we want to look for K R n with P K e i 2 = t i for all i; in short, we want to form a matrix of P K with diagonal of t i, such that P K = PK = P K 2. It is not obvious that such a thing is even possible. If we try to find a projection in that way we will get n(n+1) 2 equations with n(n 1) 2 variables, as we see in the next example. Example Take P K to be a 2 2 matrix such that P K = PK = P K 2, and such 19

26 that the diagonal of P K is (t, 1 t) for a fixed t [0, 1]. So P K = t x x, P 2 K = t2 + x 2 x. 1 t x x 2 + (1 t) 2 As these two should be equal, we get 2 equations t = t 2 + x, 1 t = x 2 + (1 t) 2 with the single unknown x. In this particular case one can check that x = 1 t 2 gives a solution. But for a matrix we will have 55 equations in 45 unknowns. The bigger the projection, the more equations we have to deal with, and the systems will always be over-determined. This is an issue, because such systems may have no solution. We will soon see, however, that this problem can be solved in general and that the Schur-Horn Theorem is the way to go. 2.2 The Schur-Horn Theorem in the Finite Dimensional Case The Schur-Horn theorem characterizes the relation between the eigenvalues and the diagonal elements of selfadjoint matrix by using majorization. Theorem (Schur 1923 [14]) If A M n (C) sa, then diag(a) λ(a), where diag(a) is the diagonal of A and λ(a) is the eigenvalue list of A. Proof. Let A = UDU, where D is diagonal matrix and U is a unitary. Then the 20

27 diagonal of A is given by a kk = h,l = l = l D hl U kh U lk λ l U kl U kl (2.7) λ l U kl 2. Define a matrix T by T kl = U kl 2. The fact that U U = UU = I implies that T is doubly stochastic. Equation 2.7 shows that diag(a) = T λ(a). By Theorem 1.10, diag(a) λ(a). Horn [6] proved in 1954 the converse of Schur s Theorem We offer a proof following ideas of Kadison [8, Theorem 6]. A very similar proof appears in Arveson- Kadison [3, Theorem 2.1], but using only results from Kadison with no acknowledgement whatsoever of the well-known results in majorization theory that we outlined in Chapter 1. The following lemma contains Kadison s key idea. Lemma Let A M n (C) with diagonal y. Let T be a T -transform. Then there exists a unitary U M n (C) such that UAU has diagonal T y. 21

28 Proof. Let A be an n n matrix. Define a unitary U by U = i j i j ξ sin θ cos θ ξ cos θ sin θ Where ξ C, with ξ = 1, and b ij ξ = b ij ξ. Then a straightforward computation shows that diag(uau ) = t. y+(1 t). y σ, Where t = sin 2 θ, σ = (i j) S n. Theorem (Horn 1954 [6]) If x, y R n, and x y, then A M n (C) sa such that diag(a) = x, λ(a) = y Proof. Let x = (x 1,, x n ), y = (y 1,, y n ) such that x y. By Theorem 1.10 x y x = (T r T 1 )y, where T 1,, T r are T transform. Let A 1 M n (R) with diagonal y and zeroes elsewhere. By Lemma 2.15, there exists a unitary V 1 such that A 2 = V 1 A 1 V 1 has diagonal T 1 y. Similarly, there exists a unitary V 2 such that A 3 = V 2 A 2 V 2 has diagonal T 2 (T 1 y) = T 2 T 1 y. Repeating this, 22

29 after r steps, we will have unitaries V 1,, V r such that A = V r V 1 A 1 V1 Vr has diagonal T r T 1 y = x. As unitary conjugation preserves the spectrum, A has spectrum y and diagonal x. We can rephrase Schur s result by saying that, that for every x R n, {M x : x y} D{UM y U : U U(n)}, where M x, M y are diagonal matrices that have x, y at the diagonal. So if we conjugate M y with the unitary matrix U we still have the same vector y at the diagonal of M y. And Horn proved the other inclusion, i.e. for every x R n, {M x : x y} D{UM y U : U U(n)}. So we can rephrase both theorems together as follows : Theorem (Schur-Horn Theorem) For every x R n, {M x : x y} = D{UM y U : U U(n)}. Now we can prove (CT-6) as follows: Proof of (CT-6). If a = (a 1,, a n ) [0, 1] n and n j=1 a j = m, then a m times {}}{ ( 1,, 1, 0,, 0). By the Schur-Horn Theorem, there exists a self adjoint matrix m times {}}{ P M n (C) with diagonal a and eigenvalues ( 1,, 1, 0,, 0). The minimal polynomial of P is f(t) = t(1 t). So P (I P ) = 0, i.e. P = P 2. Thus, P is a 23

30 projection. 2.3 A Pythagorean Theorem for Finite Doubly Stochastic Matrices In this section we consider, following Kadison [8, 9], certain differences of sums of entries of doubly stochastic matrices. We will use results here to generalize Theorem 2.18 below to the infinite dimensional case (Chapter 3). Theorem Let K be an m-dimensional subspace of H, and e 1,, e n an orthonormal basis for H. If a = r P Ke i 2, b = n i=r+1 P K e i 2, then a b = m n + r. Proof. As P K = I P K, we have P K e i 2 = P K e i, e i = (I P K )e i, e i = 1 P K e i, e i = 1 P K e i 2. So a = r a i, b = n i=r+1 1 a i, where a i = P K e i 2, and thus (using Theorem 24

31 2.9) a b = = = r a i ( 1 a i ) r a i + i=r+1 a i 1 i=r+1 r+1 a i (n r) = tr (P K ) (n r) = m n + r. Definition Let A M m,n (R). Fix subsets K {1,, m}, L {1,, n}. Then we can construct the block or submatrix B by taking only the rows in K and the columns in L of A. The complement of the block B is the matrix B with remaining rows and columns of A. The sum of all entries of the block B is the weight of the block B, and we write it as w(b). Definition A doubly stochastic matrix A = (a ij ) is said to be Pythagorean if there is a Hilbert space H with {e i }, {f j } orthonormal basis, such that a ij = e i, f j 2, for all i, j. The following result can be seen as a Pythagorean Theorem for doubly stochastic matrices. Theorem If A is a doubly stochastic matrix and B is a block in A with p rows and q columns, then w(b) w(b ) = p q + n. 25

32 Proof. w(b) w(b ) = j K l L = j K a jl j K l L a jl (1 l L a jl ) l L ( = K a jl j K l L = K L = K (n L ) = K + L n 1 j k L a jl j K l L a jl ) We can use Theorem 2.21 to give another proof of Theorem 2.18: Proof. Let K be an m-dimensional subspace of H, and K its orthogonal complement. Choose {e 1,, e n }, {f 1,, f m }, {f m+1,, f n } to be orthonormal bases for H, K and K. Let A = (a jk ) be the n n doubly stochastic matrix given by a jk = e j, f k 2. Let B be the submatrix of A given by the first r rows and m columns of A. If P k is the projection onto K, then P k e i = m j=1 e i, f j f j, (1 P k )e i = n j=m+1 e i, f j f j and P K e i 2 = m j=1 e i, f j 2 = m j=1 a ij, so r P Ke i 2 = m j=1 a ij = w(b). r Similarly, P K e i 2 = n j=m+1 e i, f j 2 = n j=m+1 a ij, so n i=r+1 P K e i 2 = n j=m+1 a ij = w(b ). By Theorem 2.21, n i=r+1 r P K e i 2 P K e i 2 = w(b) w(b ) = m n + r. i=r+1 26

33 Chapter 3 The Carpenter Theorem in the Infinite Dimensional Case In the second chapter we dealt with the finite dimensional space R n, and we showed many cases of the Pythagorean Theorem. Here we will deal with infinite dimensional Hilbert spaces and we will discuss two cases of the Carpenter Theorem. The first case when the subspace K H and its orthogonal complement K have infinite dimension, and the second case when one of the subspaces K, K has finite dimension [9]. We include here several definitions that we will need to refer to operators on an infinite-dimensional Hilbert space. Definition 3.1. For an operator A B(H) we say A is positive if A = A, and the spectrum of A consists of positive real numbers. i.e. if (Ax, x) 0, x H. Definition 3.2. Let {e i } be an orthonormal basis in B(H) and A B(H) be a 27

34 positive operator. Then we say A is a trace-class operator if tr (A) = Ae i, e i <. If the sum is finite for one orthonormal basis, then it is finite and has the same value for any other orthonormal basis. For an arbitrary A, we say it is trace-class if (A A) 1 2 is trace-class. 3.1 The Subspaces K and K both have Infinite Dimension We will start with some facts that we are going to use later. Lemma 3.3. If P B(H) is a projection that has p 1, p 2, as diagonal elements, and σ : N N is any bijection, then there exists a projection P B(H) with p σ(1), p σ(2), as diagonal elements. Proof. We have P B(H) is a projection with diagonal {p i } i {1,2, } i.e. p i = Pe i, e i, where {e i } is an orthonormal basis. We define a unitary U by Ue i e σ(i). Then we define a projection P = U PU. The i th diagonal element of this projection is given 28

35 by P e i, e i = U PU e i, e i = P Ue i, Ue i = Pe σ(i), e σ(i) = P σ(i). Lemma 3.4. Let α 1, α 2,, α n, β [0, 1], such that α 1 +α 2 + +α n = β+m, m N. Then {}}{ (α 1, α 2,, α n ) (β, 1,..., 1, 0). m Proof. Since α j [0, 1] for all j, we have α 1 1 α 1 + α α α m m. As α 1 + α α n = β + m, we have for any k {m + 1,, n}, α α k α 1 + α α n = m + β. The following lemma is proven in [5]. While the result certainly looks obvious as expected from the finite-dimensional analogue and its proof is not very hard, it is not elementary either. We will generalize it in the proof of Theorem

36 Lemma 3.5. Let P, Q B(H) be two orthogonal projections such that P Q is trace-class. Then tr (P Q) Z. Lemma 3.6. Let T B(H) be a trace class operator. Let R 1, R 2, be pairwise orthogonal finite-rank projections with k R k = I. Then tr (T ) = k tr (T R k ). Proof. k (T R k) = T k (R k) = T I. This implies T = k T R k. So, if we construct an orthonormal basis {e i } by joining orthonormal bases corresponding to each R k H, we get tr (T ) = i T e i, e i = i,k = k = k T R k e i, e i T R k e i, e i e i rang R k tr (T R k ). Let K be subspace in H and consider the orthogonal projection Q K onto K. Let q 1, q 2, be the diagonal of Q K. If Q K has finite rank (i.e. if dim K < ), then j q j = tr (Q K ) N. But what happens if j q j = (i.e. dim K = )? Since Q K is a projection, then I Q K is going to be a projection too with diagonal 1 q 1, 1 q 2,. Therefore tr (I Q K ) = j (1 q j) will have to be an integer if finite (this is an elementary exercise in Functional Analysis, or it can be obtained from Effros Lemma 3.5). For example, if q j = 1 1 j 2, then the diagonal of I Q K would be j 1 j 2 = π2 1, not an integer. So no projection with diagonal {1 } 6 j 2 30

37 exists. That is, when K is finite-dimensional or it is orthogonal complement K is, we obtain an obstruction to what the possible diagonals of Q K are. We will address this case in Section 3.2. What about when K, K are both infinite-dimensional (i.e. j q j = j (1 q j) = )? The next Theorem (3.8) will illustrate this condition and give us the complete idea of the whole situation. The proof of Theorem 3.8 in the original paper [9] is kind of complicated, so we tried to simplify it as much as we could. Definition 3.7. Let {e n } be an orthonormal basis. Then we define the matrix units {E mn } m,n associated to {e n } as the rank-one operators E mn x = x, e n e m, x H. The following is the main result in this thesis. Theorem 3.8. Given {e j } j N an orthonormal basis of H and {a j } j N [0, 1], the following statements are equivalent: 1. There exists an infinite dimensional subspace K H with infinite dimensional complement such that P k e j 2 = a j j N. 2. j N a j = and j N (1 a j) = ; and either (i) or (ii) holds: (i) a = or b = ; (ii) a <, b <, and a b Z, where a = a j 1/2 a j, b = a j >1/2 a j. Proof. 2(i) = (1): First we will show that when a =, then there exists a projection P with diagonal 31

38 {a j }. Let N 0 = {j : a j = 0 or a j = 1}, K = span{e j : j N 0 }. Then for j / N 0, a j (0, 1). If we find P 0 on B(K ) with diagonal {a j : j / N 0 }, then P = P 0 + P 1 satisfies 1, where P 1 = j N 0 a j E jj. So we will assume that a j (0, 1) for all j. We consider a decomposition {a j } = {a j } {a j}, where 0 a a j 1, so a, b will be j=1 a j, j=1 1 a j. Let a j j 1 2, and 1 2 < = a γ(j), a j = a δ(j), and N = {δ(n) : n N}, N = {γ(n) : n N}. Let n(1) = min{n : a 1+a 1+ +a n 3}; notice that each a j 1 2, a 1 < 1 so n(1) 5. Let b 1 = a 1 (b 2,, b n(1) ) = (a 2,, a n(1)). Let m(1) = min{n : a 1 + b b n 3}; then 5 m(1) n(1). Let ǎ = 3 a 1 b m(1) 1 = b m(1) 1 + ǎ b m(1) = b m(1) ǎ. m(1) 1 1 b j As a 1 + m(1) 1 m(1) 2 b j < 3, a 1 + m(1) 2 1 b j + b m(1) 1 = 3, we have ǎ 0 and 0 b m(1) < b m(1) b m(1) 1 < b m(1) 1 1. Let N 1 = {δ(1), γσ 1 (1),, γσ 1 (m(1) 1)}, where b j = a σ 1 (j) = a γσ 1 (j) for certain permutation σ 1. Let j(1) = γσ 1 (m(1)), j(2) = δ(2), and {j(n)} n 3 an increasing enumeration of N \(N 1 {j(1), j(2)}). Let n(2) = min{n : a 2 +b m(1) + n 3 a j(k) 3}. 32

39 Let c 1 = b m(1) c 2 = a 2 c 3 = a j(3) (c 4,, c n(2) ) = (a j(4),, a j(n(2)) ). Let m(2) = min{m : m j=1 c j 3}; then m(2) 1 j=1 c j < 3, 6 m(2) n(2). Define m(2) 1 ˇb = 3 j=1 c m(2) 1 = c m(2) 1 + ˇb c m(2) = c m(2) ˇb. c j Then 0 ˇb c m(2) 1, and 0 c m(2) c m(2) c m(2) 1 c m(2) 1 1. Let N 2 = {j(1), j(2)} {n : k, 3 k m(2) 1 with c k = a n }. Let k(1) be such that a k(1) is c n(2), k(2) = δ(3), write N \ (N 1 N 2 ) = {k(1), k(2), } with {k(3), k(4), } in ascending order. 33

40 Let n(3) = min{n : a 3 + c m(2) + n r=3 a k(r) 3} and let d 1 = c m(2) d 2 = a 3 d 3 = a k(3) (d 4,, d n(3) ) = (a k(4),, a k(n(3)) ), and let m(3) = min{n : m j=1 d j 3}. Then m j=1 d j 3, m(3) 1 j=1 d j 3, and 6 m(3) n(3). Let č = 3 m(3) 1 j=1 d m(3) 1 = d m(3) 1 + č, d m(3) = d m (3) č, d j, so that 0 < č d m(3) 1 2, and 0 d m(3) < d m(3) d m(3) 1 < d m(3) 1 1. By repeating these processes we will build pairwise disjoint subsets N 1, N 2, of N such that N 1 N 2 = N. We can write N j = {p j (1),, p j (m(j) 1)}, then b m(j 1) + a j + m(j) 2 k=3 a pj (k) + b m(j) 1 = 3. (3.8) By Theorem 2.12 we will get a self adjoint projection E j with diagonal {b m(j 1), a j, a pj (3),, a pj (m(j) 2), b m(j) 1}. 34

41 We also have b m(1) + b m(j) 1 = b m(1) + b m(j) 1, (3.9) and 0 b m(j) b m(j) b m(j) 1 b m(j) 1. (3.10) If we write P = E j, we get j=1 P = a 1 a p1 (2) a p1 (m(1) 2) b m(1) 1 b m(1) a 2 0 a p2 (2) b m(2) 1 b m(2).... The projection P has all the a j in its diagonal with the exception of the pairs 35

42 b m(j) 1, b m(j) in place of b m(j) 1, b m(j). We will now construct a unitary operator that will conjugate b m(j) 1, b m(j) into b m(j) 1, b m(j). So let U be U = sin θ 1 cos θ 1 cos θ 1 sin θ sin θ 2 cos θ 2 cos θ 2 sin θ 2..., where θ 1, θ 2, are to be determined. Let Q = UP U. Then every entry of Q outside the 2 2 blocks agrees with P. In the 2 2 blocks, sin θ j cos θ j cos θ j b m(1) 1 0 sin θ j sin θ j 0 b m(1) cos θ j cos θ j = sin θ j 36

43 b m(1) 1 sin2 θ + b m(1) cos2 θ b m(1) sin2 θ + b m(1) 1 cos2 θ The conditions 3.9 and 3.10 guarantee that for each j there exists t j [0, 1] such that b m(j) = t j b m(j) + (1 t)b m(j) 1, b m(j) 1 = (1 t j )b m(j) + t j b m(j) 1. Choosing θ j so that t j = sin θ j, we get the desired diagonal for Q. This proves the case a =. When b =, we can repeat the proof for the coefficients b i = 1 a i. That way we obtain a projection with E with diagonal 1 a j. Then I E is the projection we are looking for. 2(ii) = 1: We will show that if a <, b <, a b Z, then there exists a projection P with diagonal {a j }. By using Lemma 3.3, we can reorder the numbers {a j } [0, 1] as needed. Again write 0 a j 1 2 < a j 1, where {a j } {a j} = {a j }. Let a j = a γ(j), a j = a δ(j) such that N = {γ(n) : n N}, N = {δ(n) : n N}. Then a = a j N a j, b = a j N 1 a j. Since a, b are finite, we can get finite subsets N 1 N, N 1 N such that γ 1 = N \N 1 a j < 1, δ 1 = 1 a j < γ 1. N \N 1 37

44 We are given a b Z, so a b = N 1 a j + γ 1 ( N 1 (1 a j ) + δ 1 ) = a j + a j + γ 1 δ 1 N 1 N 1 N 1 = a j N 1 + γ 1 δ 1 Z. N 1 N 1 Then m = N a 1 N j + γ 1 δ 1 Z. Write N 1 N 1 = {k 1,, k r }. Since 1 0 γ 1 δ 1 < 1, 0 a j 1, and m = a k1 + + a kr + (γ 1 δ 1 ), we get from Lemma 3.4 that (a k1,, a kr, (δ 1 γ 1 )) m times r+1 m times {}}{{}}{ (1,, 1, 0,, 0). By Theorem 2.12 (CT-6), there exists a projection P 0 such that a k1... * P 0 =. a kr * γ 1 δ 1 38

45 By adding the element 1 to the diagonal of P 0 we get another projection P 1 P 1 = a k1... * 0 a kr * δ 1 γ Now we will pay attention only to the 3 3 block at the middle. Let N \ N 1 = {l 1, l 2, }, N \ N 1 = {m 1, m 2, }. We know 0 a li < 1, 0 a mi < δ 1 i N. Let γ 2 = γ 1 a l1, δ 2 = δ 1 (1 a m1 ), then γ 2 δ 2 = γ 1 δ 1 (a l1 + a m1 ) + 1 (3.11) γ 2 δ 2 + a l1 + a m1 = γ 1 δ (3.12) From Lemma 3.4, (a l1, a m1, γ 2 δ 2 ) (γ 1 δ 1, 1, 0). By the Schur-Horn Theorem 2.17 there exists a 3 3 self adjoint matrix U 1 such that 39

46 a l a m γ 2 δ 2 = diag (U 1 γ 1 δ U 1 ). So U 1 P 1 U 1 = a k a kr a l1 a m1 γ2 δ Now we define P 2 as 40

47 P 2 = a k a kr al1 a m1 γ2 δ i.e. P 2 = U 1 P 1 U 1 + E r+4,r+4. Since 0 a l2 γ 2, 0 1 a m2 δ 2, and by letting γ 3 = γ 2 a l2, δ 3 = δ 2 1 a m2, then again by Lemma 3.4, (a l2, a m2, γ 3 δ 3 ) (γ 2 δ 2, 1, 0). If we keep repeating the process we will end up with a sequence of projections {P n } B(H) such that diag(p N ) = a k1,, a kr, a l1,, a ln, a m1,, a mn, γ n+1 δ n+1, 0,. As the unitary U j+1 is the identity except possibly in the first r + 3j + 3 basis 41

48 elements, P i Q r+3j+4 = P j Q r+3j+4, i j (3.13) where is Q s = s h=1 E hh. This shows that the sequence {P n } converges strongly, Indeed, for any i j, ξ H, (P i P j )ξ = (P i P j )Iξ = (P i P j )[(I Q r+3j+4 ) + Q r+3j+4 ]ξ (P i P j )(I Q r+3j+4 )ξ + (P i P j )Q r+3j+4 ξ ( P i + P j ) (I Q r+3j+4 )ξ = 2 (I Q r+3j+4 )ξ. When h, Q h I strongly, so we find out that the sequence {P n ξ} is Cauchy in H, and so convergent. This shows {P n } B(H) is strongly convergent to a projection P. As P i Q r+3j+4 = P j Q r+3j+4, the diagonal of P is {a j }. (1) = (2): The fact that K, K are infinite-dimensional imply that j a j = j 1 a j =. So we need to show that either 2(i) or 2(ii) holds. If 2(i) holds, we are done. Otherwise we want to show that if there is a projection P B(H) such that diag(p ) = a n, then a b Z, where a = a i <, and b = (1 a i ) <. a i N a i N The proof of this is inspired by Effros proof of Lemma 3.5. Let Q B(H) be the 42

49 projection Q = n N E nn. Then tr (QP Q) = tr ( n N E nn P E nn ) = tr ( n N a n E nn ) = n N a n = a, and tr (Q P Q ) = tr ( E nn (I P )E nn ) n N = tr ( (1 a n )E nn ) n N = (1 a n ) n N = b. This implies that both QP Q, Q P Q are trace class, since they are positive. We now notice that P Q is Hilbert-Schmidt, and so in particular it is compact. 43

50 Indeed, using that P = P 2, (1 P ) 2 = (1 P ), we have tr ((P Q ) 2 ) = [(P Q )] hh h = P hk Q hk 2 (no problem exchanging the sums h k = P hk Q hk 2 since every term is non-negative) k h = P hk 2 + (1 P kk ) 2 + P hk 2 k N h k N n k = P kk + (1 P kk ) k N k N = a + b <. Since (P Q ) 2 is positive and compact, then we can write (P Q ) 2 = k λ kr k, where R k = R 1, R 2, are pairwise orthogonal finite rank projections and {λ i } i N arranged strictly in decreasing order and converging to zero. The points {λ i } are isolated points in the spectrum of (P Q ) 2, so there exist continuous functions f 1, f 2, such that E k = f k ((P Q ) 2 ). Since P (P Q ) 2 = (P Q ) 2 P, Q(P Q ) 2 = (P Q ) 2 Q (from a direct computation), we deduce that P R k = R k P, QR k = R k Q, k; in particular P R k, QR k are both finite rank projections. By keeping in 44

51 mind that QP Q, Q P Q are trace class and using Lemma 3.6, a b = tr (QP Q) tr (Q P Q ) = tr (QP Q Q P Q ) = k = k = k = k = k = k tr [(QP Q Q P Q )R k ] tr (QP QR k ) tr (Q P Q R k ) tr (QP R k QR k ) tr (Q P R k Q R k ) tr (P R k QR k P R k Q R k ) tr (P R k QR k (R k R k QR k P R k + P R k QR k )) tr (P R k + QR k R k ). P R k, QR k, and R k are finite rank projections and their traces are integers, so tr (P R k + QR k R k ) Z, k. As every term in the series is an integer, we conclude that they are eventually zero, and that a b Z. 3.2 One of the Subspaces K, K has Finite Dimension In this context we will show the Carpenter s theorem for the finite dimensional subspaces K, K of an infinite dimensional Hilbert space, where the sum of the diagonal elements of the projection P k is finite and integer. The proof comes as consequence of Theorem

52 Theorem 3.9. If {e j } j J is an orthonormal basis for an infinite-dimensional Hilbert space H, and {t j } j J [0, 1], then the following statements are equivalent: 1. There exists m-dimensional K H, such that P k e j 2 = t j. 2. j J t j = m. Proof. First we will show (2) (1): As in Theorem 3.8, we write {t j } = {t j} {t j }, 0 t j 1, and 1 < 2 2 t j j t2 j <, t j 0, so {t j } is necessarily finite, i.e. {t j } = {t 1,, t k }. Let 1. As a = t j, b = 1 1 (1 t j ). Then a b = = t j 1 t j k 1 1 = m k Z. (1 t j ) Since a b Z we apply Theorem 3.8 to get a projection P K such that the diagonal of P K is t 1, t 2,. And dimk = tr (P K ) = j=1 P Ke j, e j = j=1 t j = m. For the other direction (1) (2): Let P K be the orthogonal projection onto K. As dim K = m, we get m = tr (P K ) = j P K e j, e j = j P K e j 2 = j t j. When K has finite co-dimension, we can apply Theorem 3.9 to its orthogonal 46

53 complement K to obtain a projection P K with prescribed diagonal {s j } and P K = I P K. So the diagonal of P K is {1 s j }. We get thus the following result: Theorem If {e j } is an orthonormal basis of H, and {t j } j J [0, 1], then the following statements are equivalent: 1. There exists K H, of co-dimension m, such that P k e j 2 = {t j } j J. 2. j J (1 t j) = m. 3.3 A Pythagorean Theorem for Infinite Doubly Stochastic Matrices In chapter 2 we defined Pythagorean matrix (2.20), then we studied finite doubly stochastic matrices A, and we calculated the weight of some block B on the matrix A and the weight of it complementary block. Then we proved a Pythagorean Theorem (Theorem 2.21) for finite doubly stochastic matrices, where w(b) w(b ) Z. For an infinite doubly stochastic matrix we can t do that immediately because we are dealing with infinite rows and columns. Kadison defines the weight for an infinite doubly stochastic matrix, by taking the sum as a limit. The sets of the rows and the columns are infinite sets, so he takes all the family of the finite subsets, that ordered by inclusion gives us a net, and then he takes the limit over the net. By using Theorem 3.8 we will show a Pythagorean Theorem for infinite Pythagorean matrix when it has infinite complementary blocks, each of them with finite weight. Definition We say that a doubly stochastic matrix A is orthostochastic if A = U ij 2, where U is unitary matrix [4]. 47

54 Lemma If A is a Pythagorean matrix, then it is doubly stochastic. Proof. To show that a Pythagorean matrix is doubly stochastic we will show that it is orthostochastic. Let e i, f j be orthonormal bases with A = e i, f j 2. We know that there exists a unitary U, with Uf i = e i. So e i, f j 2 = Uf i, f j 2 = U ji 2. How can an infinite doubly stochastic matrix that is Pythagorean have infinite complementary blocks with finite weights? To make that clear we will discuss an example. Let 2 i if i Z a i = (1 2 i ) if i Z + and let a = i Z a i = 1, b = i Z + 1 a i = 1. Write Z 0 = Z + Z. By using Theorem 3.8, since a b Z, there exists an infinite dimensional K H, with infinite dimensional orthogonal complement K such that the diagonal of the projection P K is {a i } i Z0. When j Z +, (I P K )e j 2 = 1 P K e j 2 = 1 a j = 2 j. Let {f j } j Z+, {f j } j Z be orthonormal bases for K, K respectively. Let a ij = e i, f j 2 ; then A = (a ij ) is an infinite Pythagorean matrix. a ij = e i, f j 2 = P K e i 2 = a i, i Z j Z + j Z + and so j Z a ij = (I P k )e i 2 = 1 a i, i, 48

55 a ij = a i = 1 j Z + i Z i Z and a ij = a j = 1. i Z j Z + j Z + Thus the weight of both complementary blocks is finite, and the difference is an integer. Theorem If A is an infinite Pythagorean matrix, and B, B are a block and its complement in A, and both their weights are finite, then w(b) w(b ) Z. Proof. Since both blocks have finite weight, then both B, B are infinite (because the complement of a finite block has infinite weight). Let us write A = (a ij ), i, j Z 0, where B = (a ij ) i,j Z, and B = (a ij ) i,j Z+. As A is Pythagorean, there exist {e i } i Z0, {f j } j Z0 orthonormal bases for H such that e i, f j 2 = a ij i, j Z 0. Let K H be the subspace that is spanned by {f j } j Z. Then w(b) = a ij = e i, f j 2 i,j Z i Z j Z = e i, f j f j, e i i Z j Z = P K e i, e i = P K e i 2. i Z i Z Similarly, w(b ) = a ij = e i, f j 2 = (I P K )e i 2 = 1 P K e i 2. i,j Z + j Z + j Z + i Z + i Z + i Z + 49

56 As both weights are finite, (i) (ii) in Theorem 3.8 gives us w(b) w(b ) Z. If we compare the finite and infinite dimensional versions of the Pythagorean Theorem for doubly stochastic matrices 2.21 and 3.13, we note that in the finitedimensional case the result holds for any doubly stochastic matrix, while the infinite dimensional case requires an additional hypothesis (i.e. Pythagorean). We don t know if Theorem 3.13 holds for arbitrary doubly stochastic infinite matrices. We note below that not every doubly stochastic matrix is Pythagorean, starting with dimension 3. For example if we take any 2 2 doubly stochastic matrix, A = a 1 a 1 a a and then construct 2 orthonormal bases e 1, e 2, and f 1, f 2, where f 1 = a e a e 2, f 2 = 1 a e 1 + a e 2, then A ij = e i, f j 2, so A is Pythagorean. What about 3 3 doubly stochastic matrices? Consider A = If A were Pythagorean, n there would exist orthonormal bases {e 1, e 2, e 3 }, {f 1, f 2, f 3 } 50

57 with A ij = e i, f j 2. Then 0 = A 31 = e 3, f 1 2, so f 1 = se 1 + re 2, 0 = A 22 = e 2, f 2 2, so f 2 = pe 1 + qe 3. Also, 1 2 = A 11 = e 1, f 1 2 = s 2, so s = A 12 = e 1, f 2 2 = p 2, so p 0. But then f 1, f 2 = sp 0, contradicting the fact that {f 1, f 2, f 3 } is an orthonormal basis. So every 2 2 doubly stochastic matrix is Pythagorean, but bigger doubly stochastic matrices will not necessarily be Pythagorean. 51

58 Chapter 4 A Schur-Horn Theorem in Infinite Dimensional Case In this chapter we will see a generalization for Schur-Horn Theorem in infinite dimension by A. Neumann [11]. After that we will give an explanation for this result. After that, we will consider W. Arveson and R. Kadison s Schur-Horn theorem for positive trace-class operators on an infinite dimensional Hilbert space [3]. 4.1 Majorization in Infinite Dimension To begin, first we will define majorization in infinite dimension and for that we will use l (N) B(H) instead of using R n, such that l (N) has real entries. We chose l (N) B(H) here because the diagonal of a self adjoint operator as a matrix is a bounded sequence of real numbers, so we can think here about l (N) as diagonal self adjoint matrices inside B(H). We defined majorization in finite dimension as follows: 52

59 For any two finite vectors x, y R n, we say x is majorized by y, denoted x y, if and x i = x i y i for k < n, y i. One can try to generalize this, naively, by saying that for x, y l (N), x is majorized by y if and x i = x i y i for k <, y i. But such majorization would fail to generalize many of the properties that finitedimensional majorization enjoys. For instance in finite dimension (Proposition 1.4), we have that x y if and only if and x i = y i x i k < n, y i. But in l (N), the numbers x n are not defined if x = 1 1, and neither are n x n if x n = 1. Also, looking at Theorem 1.10, the sequences x = (1, 1, 1, ) and n 53

60 y = (2, 2, 2, ) would satisfy x y, but x / convpy (not even in the closure). What A. Neumann does to solve this problem, instead of taking these sums y i x i k < n, he defines U k (x) = sup{ i F x i : F = k} i.e. he takes all possible sums of cardinality k in x, and then he takes the supremum. But this is not enough, we can see the reason in the following example. Example 4.1. Let x, y l (N) such that x = (1, 1, 1, 1, ) and y = (1, 0, 1, 0, ), then U k (x) = k and U k (y) = k. So both vectors would majorize each other, and having double majorization one would expect both vectors to be permutations of each other, as in Theorem 1.3. So we would have x y, y x x = Py, which in this example is clear that can t happen. Note that if we have x i y i, and y i x i k 54

61 then x i = y i. Neumann uses this idea to (implicitly) define majorization in infinite dimension as follows [11]: Definition 4.2. Let x, y l (N), then we say x y if k N, U k (x) U k (y) and L k (x) L k (y), where U k (x) = sup{ i F x i : F = k}, L k (x) = inf{ i F x i : F = k}. So, in Example 4.1, U k (x) = L k (x) = k, U k (y) = k, L k (y) = 0; then x y is true, but y x. 4.2 Neumann s Schur-Horn Theorem We start by defining some terminology to be used later [15]. Definition 4.3. A subalgebra A B(H) is maximal abelian if any two elements from A commute, and A is not properly contained in any other commutative subalgebra A B(H). Definition 4.4. A von Neumann algebra M on H is a -subalgebra of B(H) such that M = M, where M is the double commutant of M. 55

62 Definition 4.5. We say that a von Neumann M is atomic when every nonzero projection majorizes a nonzero minimal projection. Definition 4.6. Let A M be subalgebra a von Neumann algebra M. Then we say a linear map E is a conditional expectation when E : M A is onto, E = E 2, and E = 1. Moreover, we say that E is trace preserving when tr E = tr. we have In the finite dimensional case, Schur-Horn Theorem 2.17 states for every x, y R n {M x x : x y} = D{UM y U : U U(H)}. If we want to do the analog thing for infinite dimensional case, take x, y l (N), where we see l (N) as the diagonal operators for a fixed orthonormal basis. If we write E for the projection onto the diagonal, then we expect {M x : x y} = E{UM y U : U U(H)}. This equality cannot hold without norm closure: if we go back to our Example 4.1, were x = (1, 1, 1, 1, ) and y = (1, 0, 1, 0, ), then x y but M x = I / E{UM y U : U U(H)}, while M x = I E{UM y U : U U(H)}. So the closure seems to be necessary to satisfy the equality of Schur-Horn Theorem in the infinite dimensional case, and this is what Neumann does to form the next theorem ([11], Corollary 2.18, and Theorem 3.13). 56

63 Theorem 4.7 ([11]). For x, y l (N) we have {M x : x y} = E{UM y U : U U(H)}. 4.3 A Strict Schur-Horn Theorem for Positive Trace-Class Operators We will define positive trace-class operators and then we will form a Schur-Horn Theorem for positive trace-class operator on infinite dimensional Hilbert space. We call this theorem strict, because no closure is required after projecting onto the diagonal. We refer to the definitions of positive and trace-class operators at the beginning of Chapter 3. Definition 4.8. For a trace-class operator A, its one-norm is A 1 = tr ((A A) 1 2 ). Definition 4.9. Let A, B B(H) be two trace-class operators. We say A and B are L 1 equivalent if there exist a sequence of unitary operators {U i } such that A U n BU n 1 0, when n. 57