Math 408 Advanced Linear Algebra

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1 Math 408 Advanced Linear Algebra Chi-Kwong Li Chapter 4 Hermitian and symmetric matrices Basic properties Theorem Let A M n. The following are equivalent. Remark (a) A is Hermitian, i.e., A = A. (b) x Ax R for all x C n. (c) A is normal with real eigenvalues. (a) Let A M n. Then A = H + ig, where H = (A + A )/2 and G = (A A )/(2i) are Hermitian. Also, AA and A A are Hermitian. If A is Hermitian, then A k is Hermitian for positive integer k, and A 1 is Hermitian if A is invertible. (b) The set of Hermitian matrices form a real linear space. (c) The product of Hermitian matrices may not be Hermitian. (d) The product of two Hermitian matrices is Hermitian if and only if they commute. (e) A matrix A is the product of two Hermitian matrices if and only if A is similar to A (via a Hermitian matrix), equivalently, A is similar to a real matrix. Proof. (a) If A is Hermitian, then A = U DU for some real diagonal matrix D. So, A k = U D k U is Hermitian for positive integers k (or negative integers if A 1 exists). (b) Let H n be the set of n n Hermitian matrices. Then 0 H n, and ra + sb H n for any A, B H n and r, s R. (c) Consider A = E 12 + E 21 and B = E 11 E 22. (d) If A, B H n commute then A = U D 1 U and B = U D 2 U for some unitary U and real diagonal D 1, D 2. Thus, AB = U (D 1 D 2 )U is unitary. If A, B, AB H n, then AB = (AB) = B A = BA. (e) Suppose A = HK with H, K H n. If H = U DU for some unitary U and D = D 1 0 n k, ( ) where D 1 is an invertible real diagonal matrix in H k. Then U AU = DU D1 K KU = and U A U = ( K11 D ). Now, K 11 D 1 and DK 11 have the same Jordan blocks for the nonzero eigenvalues, and rank (A m ) = rank ((A ) m ) for all m. Thus, A and A have the same Jordan form and are similar. Suppose AS = SA with S = H+iG. Then S A = AS implies that A(H+iG) = (H+iG)A and (H ig)a = A(H ig). Hence, AH = HA and AG = GA. Hence AK = KA for some invertible K = H + tg. Thus, A = K(A K 1 ), where (A K 1 ) = K 1 A = A K 1. Suppose A and A are similar. Then they have the same Jordan blocks. Thus, the Jordan blocks J m (λ) and J m (λ) will occur together in A for any nonreal eigenvalue λ, and can be combined to a real Jordan block. Thus, A is similar to a direct sum of real Jordan blocks. 1

2 Suppose S 1 AS = R for a real matrix R. Then S 1 AS = R = T 1 R t T = T 1 R T = T 1 S A (S 1 ) T. Thus, A and A are similar. Eigenvalue inequalities Denote by λ 1 (A) λ n (A) the eigenvalues of a Hermitian matrix A M n. Theorem (Courant-Fischer-Weyl) Suppose A M n is Hermitian. Then for k {1,..., n}. λ k (A) = max dim W =k min v W,v v=1 v Av = min dim W =n k+1 min v W,v v=1 v Av. Proof. For simplicity, write λ j = λ j (A) for all j. Suppose {u 1,..., u n } is a set of orthonormal eigenvectors of A such that Au j = λ j u j. For any subspace of dimension W, say, spanned by w 1,..., w k, it contains a unit vector v in the span of {u k,..., u n }. Thus, v = n j=k µ ju j with n j=k µ j 2 = 1 such that v Av = µ j 2 λ j j=k µ j 2 λ k = λ k. Let W be the span of u 1,..., u k. For any v W, we have v = k γ jv j so that v Av = k γ j 2 λ j j=k γ j 2 λ k = λ k and the equality holds if v = u k. The first equality holds. The proof is complete. Corollary (Rayleigh principle) Let A M n be Hermitian. For any unit vector x C n, j=k λ 1 (A) v Av λ n (A). Corollary (Interlacing inequalities) Let A M n be Hermitian with eigenvalues a 1 a n and B M n 1 be a principal submatrix of A with eigenvalues b 1 b n 1. Then a 1 b 1 a 2 a n 1 b n 1 a n. Theorem (Fan-Pall) Suppose a 1,..., a n and b 1,..., b n 1 satisfy the interlacing inequalities. Then there is a Hermitian matrix A M n with a principal submatrix B M n 1 with a 1,..., a n and b 1,..., b n 1 as eigenvalues. Proof. By the discussion in class, We may assume that a 1 > b 1 > a 2 > a n 1 > b n 1 > b n. Let D = diag (a 1,..., a n ). We show that there is a unitary U such that if we delete the first row and first column of the matrix A = UDU, we get a matrix B H n 1 with eigenvalues b 1 b n. To see this, note that for t / {a 1,..., a n }, we have ti A = U(tI D)U is invertible with inverse (ti A) 1 = U(t D) 1 U. (1) 2

3 Recall that the (1, 1) entry of an invertible matrix C equals det(c(1, 1))/ det(c) using the adjoint matrix formula, where C(1, 1) is obtained from C by removing its first row and first column. So, the (1, 1) entry of (ti A) 1 equals det(ti B)/ det(ti A) as B is the matrix obtained from A by removing the first row and the first column. Now, consider the right side of (1). If the first row of U equals (x 1,..., x n ), then the (1, 1) entry of the matrix is f(t) = x 1 2 t a x n 2 t a n. Equating the two expressions for the (1, 1) entry of (ti A) 1 mentioned above, we have det(ti B) = x j 2 t a j det(ti A) = x j 2 a k ). k j(t Homework 1. (a) Show that for all i = 1,..., n, the value w i = n 1 (a i b j )/ j i (a i a j ) is positive. By our assumption that a 1 > b 1 > b n 1 > b n, we see that for j i, we have a i > b j if and only if a i > a j. Thus, the numerator and the denominators have the same number of negative terms. Thus, the quotient is positive. (b) If we choose x i = w i for all i = 1,..., n, show that x j 2 n 1 det(ti A) = (t b j ) t a j for all t = a 1,..., a n, and hence the two polynomials in t are identical. If we substitute a i on the left hand side, all the summands are zero except for the one involving x i 2, and by the definition of x i 2 the expression reduces to n 1 (a i b j ). Since both side are polynomials of degree n 1, we see that the two polynomail are identical. (c) Show that if we choose a unitary U with the first row x = (x 1,..., x n ) satisfying (b), then U AU has a principal submatrix with eigenvalues b 1 b n 1 as asserted. By part (a) and (b), and the previous discussion, such a choice of U will yield a submatrix B of A = UDU such that det(ti B) = n 1 (t b j). Thus, B has the desired eigenvalues. Definition Let c = (c 1,..., c n ) and d = (d 1,..., d n ) be real vectors. We say that c is majorized by d, denoted by c d if the sum of the k largest entries of c is not larger than that of d for k = {1,..., n 1}, and the sum of all entries of c is the same as that of d. Theorem Suppose a = (a 1,..., a n ) and d = (d 1,..., d n ) are real vectors. Then there is a Hermitian matrix in M n with eigenvalues a 1,..., a n and diagonal entries d 1,..., d n if and only if d a. 3

4 Homework 2. (a) Suppose n = 2. If (d 1, d 2 ) (a 1, a 2 ), show that there is a vector v R 2 such that v t Dv = d 1, where D = diag (a 1, a 2 ). Furthermore, if P is an orthogonal matrix with the first column equal to v, then P t DP has diagonal entries d 1, d 2. Note that a 2 d 1 a 1. Assume v t = (cos θ, sin θ). Then f(θ) = v t Dv = a 1 cos 2 θ + a 2 sin 2 θ. Now, f(0) = a 1 d 1 and f(π/2) = a 2 d 1. By continuity, there is θ [0, π/2] such that f(θ) = d 1. Let w t = ( sin θ, cos θ). Then P = [v w] is orthogonal, P t DP has (1, 1) entry d 1. By the trace condition, we see that the (2, 2) entry of P t DP is d 2 so that a 1 + a 2 = d 1 + d 2. (b) Suppose n 3. Suppose (d 1,..., d n ) (a 1,..., a n ), where a 1 a n and d 1 d n. Let k be the largest integer such that a k d 1. (b.1) If a 1 a n, show that k < n and a k d 1 > a k+1. Assume a 1 a n d 1 d n. Then a a n = d d n implies that a 1 = = a n = d 1 = = d n, which is a contradiction. (b.2) Show that d = (d 2,..., d n ) is majorized by ã = (a 1,..., a k 1, ã k+1, a k+2..., a n ) where ã k+1 = a k + a k+1 d 1. Consider the sum of the l largest entries of the second vector. Case 1. If l < k, then l a j l d j l d j+1, which is the sum of the l largest entries of the first vector. Case 2. If l k, then the sum of the l largest entries of ã is at least l+1 a j d 1 which is larger than or equal to l d j+1, which is the sum of the l largest entries of d, by the original assumption. (b.3) (Optional) Show by induction that one can construct a real symmetric matrix with eigenvalues a 1,..., a n and diagonal entries d 1,..., d n. We prove by induction on n. If n = 2, the result follows from (a). Suppose n > 2. Construct the vector d and ã in (b). There is unitary (real orthogonal) V M n 1 so that V DV has diagonal entries d 2,..., d n, where D = diag (ã k+1, a 1,..., a k 1, a k+2,..., a n ). Now, there is a unitary (real ( ) ( ) orthogonal) P M 2 such that P ak 0 d1 P =. Let Q = (P I a k+1 0 ã n 2 )([1] V ). k+1 Then Q diag (a k, a k+1, a 1,..., a k 1, a k+2,..., a n )Q has diagonal entries d 1,..., d n. Theorem (Lidskii) Let A, B M n be Hermitian with eigenvalues a 1 a n and b 1 b n, Suppose C = A + B has eigenvalues c 1 c n. Then for any 1 i 1 < < i m n, m b n j+1 s=1 m (c is a is ) s=1 m b j. s=1 Corollary (Weyl) Let A, B M n be Hermitian. Then for any j, k {1,..., n} with j + k 1 n, we have λ j (A) + λ k (B) λ j+k 1 (C). 4

5 Proof. Suppose A has eigenvalues a 1 a n and B has eigenvalues b 1 b n. We may replace (A, B) by (A a j I, B b k I) and assume that a j = 0 = b k. Let A = UD 1 U and B = V D 2 V be such that U, V M n are unitary, D 1 = diag (a 1,..., a n ) and D 2 = diag (b 1,..., b n ). Set A 1 = Udiag (a 1,..., a j 1, 0,..., 0)U and B 1 = V diag (b 1,..., b k 1, 0,..., 0)V. Note that A 1 +B 1 has rank at most j +k 2, i.e., A 1 +B 1 has at most j +k 2 nonzero eigenvalues. Thus, λ j+k 1 (A 1 + B 1 ) = 0 and λ j+k 1 (A + B) λ 1 (A + B A 1 B 1 ). Now, set X = A 1 + B 1, Y = A + B A 1 B 1, and Z = X + Y = A + B. By Lidskii Theorem, we have λ j+k 1 (Z) λ j+k 1 (X) λ 1 (Y ). Next, set X = A A 1, Y = B B 1, and Z = X + Y. By Lidskii Theorem again, λ 1 (Z) λ 1 (X) + λ 1 (Y ). Thus, λ 1 (A + B A 1 B 1 ) λ 1 (A A 1 ) + λ 1 (B B 1 ) = λ j (A) + λ k (B). Positive semi-definite matrices Definition A matrix A M n is positive definite (positive semi-definite) if x Ax is positive (nonnegative) for all nonzero x C n. Theorem Let A M n. The following are equivalent. (a) A is positive definite (semi-definite). (b) A is normal with positive (nonnegative) eigenvalues. (c) A = RR, where R M n with det(r) > 0 and can be chosen to be in triangular form or positive definite (R M n with det(r) 0 and can be positive demi-definite). (d) A is Hermitian with det(a k ) > 0 for k = 1,..., n, where A k is the leading k k principal submatrix. (There is a permutation matrix P such that P AP t = Ã 0 n m such that det(ãj) > 0 for j = 1,..., m.) Proof. The equivalence of (a) and (b) are shown in class. To prove (b) (c). Let A = UDU where U is unitary, and D = diag (a 1,..., a n ). Then for D = diag ( a 1,..., a n ), we have A = RR with R = U DU. Evidently, R is positive definite. Further, we can write R = LV for some unitary V and lower triangular L so that A = LL. To prove (c) (a). Note that for any nonzero x C n, y = R x is nonzero as R is invertible. Thus, x Ax = y y > 0. Thus, A is positive definite. To prove (c) (d) if we choose R to be lower triangular in condition (c). Then det(a k ) = det(r k R k ) = det(r k) det(r k ) = det(r k) 2 > 0. To prove (d) (c), by Gaussian elimination, we can choose L 1 by changing the first column of I n so that L 1 A has zero entries in the (i, 1) positions for i = 2,..., n. Since A is Hermitian and the first row of L 1 A is the same as A, we have L 1 AL 1 = [a 11 ] A 2 where a 11 is the (1, 1) entry of A. Now, change L 1 to L 1 = L 1 ( diag ( a11, 1,..., 1) ). Then L 1 A L 1 = [1] A 2. Now, use induction argumet on A 2, we see that A 2 = L 2 L 2 for some lower triangular matrix with positive 1 diagonal entries. Set R = L 1 ([1] L 2). Then A = RR. 5

6 Definition Let A M n. Then AA and A A has the same collection of nonnegative eigenvalues. The numbers s j (A) = λ j (AA ) is the jth singular value of A for j = 1,..., n. Theorem Let A M n. (a) (SVD decomposition) There is are unitary matrices such that U AV is a diagonal matrix with diagonal entries s 1 (A),..., s n (A). (b) (Polar decomposition) There are positive semi-definite matrices P, Q and unitary matrices Ṽ, Ũ such that A = P Ṽ = ŨQ. Theorem Let A M n. Then the matrix ( ) 0 A A has eigenvalues ±s 0 1 (A),..., ±s n (A). Remark We can deduce singular value inequalities using the above (Wielandt) matrix. Congruence Theorem Let A M n be Hermitian, and S M n be invertible. Then S AS and A have the same number of positive and negative eigenvalues. In particular, there is an invertible T M n such that T AT = I p I q 0 s. Theorem Let A, B M n be Hermitian. Then there is an invertible S M n such that both S AS and S BS are in diagonal form if any one of the following is true. (a) There are real numbers a, b such that aa + bb is positive definite. (b) The matrix A is invertible and A 1 B is similar to a real diagonal matrix. Symmetric matrices For real symmetric matrices, the theory of Hermitian matrices applies. For complex symmetric matrices, we have the following results. Theorem Let A M n. The following are equivalent. (a) A is symmetric. (b) There is a unitary U M n such that A = UDU t such that D has nonnegative entries arranged in descending order. (c) A = SS t for some S M n. Theorem Let A M n. Then A is normal and symmetric if and only if there is a real orthogonal matrix A such that QAQ t is a diagonal matrix. Theorem Every matrix A is similar to a symmetric matrix. Consequently, every matrix A is a product of two symmetric matrices. Theorem Let A M n be symmetric. Then A is diagonalizable if and only if there is a complex orthogonal matrix Q such that Q t AQ is in diagonal form. 6