Part 2 Submatrices, Sum and Product of Matrices

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1 Part Submatrices, Sum and Product of Matrices In perturbation theory, one often study the change of eigenvalues and singular values of sum and product of matrices. Here we present some basic results and techniques. Sum of Matrices Theorem. (Lidskii Let A, B H n has eigenvalues a a n and b b n, respectively. Suppose C = A + B has eigenvalues c c n. For any i < < i m, b n j+ (c is a is b s. More generally, we have the following. Theorem. (Thompson Let A, B H n have eigenvlues a a n and b b n, respectively. Suppose C = A + B has eigenvalues c c n. If i < < i m and j < < j m n, then (a is + b js c is+j s s. Theorem.3 Let A M n have singular values s s n. Then ( 0n A Ã = H n A have eigenvalues ±s,..., ±s n. Theorem.4 Let A, B M n have singular values a a n and b b n, respectively. Suppose C = A + B has singular values c c n. Then for any i < < i m and j < < j m n, 0 n (a is + b js c is+j s s.

2 Product of Matrices Theorem. Let A, B H n have eigenvalues a a n and b b n. (a There exists an invertible S such that B = S AS if and only if A and B have the same inertia. (b If there exists S M n with λ n (S S such that B = S AS, then a j b j for all j =,..., n. Theorem. Let S M n be invertible and A H n. Then à = S AS and A have the same inertia. If i < < i m n are such that λ ij (A 0 or / and λ ij (à 0, then λ n j+ (S S [λ (Ã/λ ij i j (A] λ j (S S. Theorem.3 Let A, B M n have singular values a a n and b b n, respectively. Suppose C = AB has singular values c c n. Then for any i < < i m and j < < j m n, 3 Submatrices (a is b js c is+j s s. Theorem 3. Let A H n and U,..., U k M n be unitary. Then ( k k λ Uj AU j λ(a. Corollary 3. If A = (A ij i,j k H n, then λ(a A kk λ(a. Lemma 3.3 Let A and B be m n matrices. Then AB and BA have the same nonzero eigenvalues.

3 Theorem 3.4 Suppose C = (C ij i,j H n has eigenvalues c c n, C H k has eigenvalues a a k, and C H n k has eigenvalues b b n k. Set a i = c n for i {k +,..., n} and b j = c n for j {n k +,..., n}. Then for any i < < i m n and j < < j m n, [(a is c n + (b js c n ] (c is+js c n. Remark 3.5 We can obtain inequalities relating the singular values of C and the eigenvalues of C using the fact that ( 0k C = C (I k I n k C(I k I n k. C 0 n k 4 Cartesian decomposition Theorem 4. Let A, B H n have singular values a a n and b b n. If A + ib has eigenvalues z,..., z n, then Re(z,..., z n (a b n,..., a n b. Theorem 4. Let A, B H n have singular values a a n and b b n. Suppose A + ib has singular values s,..., s n. Then (a + b n,..., a n + b (s,..., s n and (s + s n,..., s n + s (a + b,..., a n + b n. Proposition 4.3 Let A M n have singular values s s n. For k {,..., n}, { k n } s j = min s j (B + ks (C : B, C M n, A = B + C. Theorem 4.4 Using the notation in Theorem 4., we have (s,..., s n w ( a + ib,..., a n + ib n w (s,..., s n. Theorem 4.5 Suppose A, B H n have singular values a a n and b b n. Then det(a + ib If A and B are positive definite, then s j (A + ib j=k a j + ib n j+. (a j + ib j. j=k 3

4 5 Tensor products Definition 5. Let A = (a ij and B = (b ij be matrices. The tensor product of A and B is the matrix A B = (a ij B. If A and B are of the same size, then the Schur product of A and B is the matrix A B = (a ij b ij, which is a submatrix of A B. Theorem 5. Let A M n and B M m have eigenvalues (respectively, singular values a,..., a n and b,..., b m. Then A B have eigenvalues (respectively, singular values a i b j with (i, j {,..., m} {,..., n}. Corollary 5.3 If A, B H n are positive definite, then λ n (A B λ n (Aλ n (B. Theorem 5.4 Suppose A M r and B M s have eigenvalues a,..., a r and b,..., b s. Then A I s + I r B have eigenvalues a i + b j with (i, j {,..., r} {,..., s}. Corollary 5.5 Suppose p(z and q(z are the integral polynomials with the algebraic numbers a and b as zeros. Let A M r and B M s be the companion matrices for p(z and q(z. Then ab is a zero of A B, and a + b is a zero of A I s + I r B. 6 Additional results and open problems The necessary and sufficient condition has been determined for the existence of Hermitian A, B and C = A + B with eigenvalues a a n, b b n, and c c n, respectively. The condition is described in term of the equality n (a j + b j = n c j and inequalities of the form a r + b s c t r R s S t T for a collection of subsets R, S, T of {,..., n}. There are similar results on (a the relations of the singular values of sum and product of matrices. (b the relations of the eigenvalues and singular values of submatrices and the entire matrix. (c the relations of singular values of A and the eigenvalues of A = (A ij i,j H n. (d the relations of the diagonal entries, eigenvalues, and singular values of sum and product of matrices. 4

5 There are additional results concerning the determinant, the rank, the eigenvalues, the inertia and the norms of sum of matrices from unitary orbits. There are many problems under current research.. Determine the complete set of eigenvalues (respectively, singular values, inertia values, ranks and norm values of U AU + V BV for unitary U, V M n.. More generally, one may consider the above problems for square matrices A M n and B M m, and use partial isometries U and V such that U U = I k and V V = I k. If A and B are adjacency matrices of two graphs, then the above problem is related to finding similar subgraphs in the two given graphs. 3. Let A = (a ij M n be real symmetric. Determine orthogonal matrices Q,..., Q n M n such that n diag (a,..., a nn = n Q t jaq j. Note that ( n A = n D j AD j, where D,..., D n are all diagonal orthogonal matrices with (, entry equal to. 4. Determine the condition on C, C 3, C 3 for the existence of C = (C ij i,j 3 H n with prescribed eigenvalues c c n. 5. Riemann hypothesis can be formulated as a problem of estimating the determinant. Let D n = (d ij M n be the divisor matrix defined by d ij = if j is a multiple of i, and d ij = 0 otherwise. Let L n M n be the matrix with at the (i, entry for i =,..., n, and all other entries equal to 0. If A n = D n + L n, then det(a n = µ(j is the Mertens function, i.e., µ(j is the Möbius function. Hence, Riemann hypothesis is true if and only if det(a n = O(n /+ε for every ε > 0. 5

6 7 Exercises. Fill in the many missing details in our discussion.. Let A, B M n (C have singular values a a n and b b n. Show that (a j + b n j+ det(a + B { 0 if [an, a ] [b n, b ], n a j b n j+ otherwise. ( R 0 3. Let A = has singular values s S T s n > 0, where R, S, T M n. Show that ( s(t SR w s n s,..., s n+ s n, and s(sr w ( s s n s n s,..., s n s n+ s n+ s n, ( s(t s S w s n s n,..., s n+. s n s s n+ s n ( A A 4. Let A, A, A, A M n and A = H A A n have eigenvalues a a n > 0. Show that s(a / A w ( a a n,..., a n a n+, and ( s(a A a an an w,..., a n a a n+ an+ a n. 6