- 4 3 Eigenvalue Inequalities For Hermitian Matrices The Following Theorem Of Hermann Weyl Is The Source Of A Great Many I 1 (257.6 KiB) Viewed 43 times
4.3 Eigenvalue inequalities for Hermitian matrices The following theorem of Hermann Weyl is the source of a great many inequalities involving either a sum of two Hermitian matrices or bordered Hermitian matrices. Theorem 4.3.1 (Weyl). Let A,B∈Mn be Hermitian and let the respective eigenvalues of A,B, and A+B be {λi(A)}i=1n,{λi(B)}i=1n, and {λi(A+B)}i=1n, each algebraically ordered as in (4.2.1). Then λi(A+B)≤λi+j(A)+λn−j(B),j=0,1,…,n−i for each i=1,…,n, with equality for some pair i,j if and only if there is a nonzero vector x such that Ax=λi+j(A)x,Bx=λn−j(B)x, and (A+B)x=λi(A+B)x. Also, λi−j+1(A)+λj(B)≤λi(A+B),j=1,…,i for each i=1,…,n, with equality for some pair i,j if and only if there is a nonzero vector x such that Ax=λi−j+1(A)x,Bx=λj(B)x, and (A+B)x=λi(A+B)x. If A and B have no common eigenvector, then every inequality in (4.3.2a,b) is a strict inequality. Proof. Let x1,…,xn,y1,…,yn, and z1,…,zn be orthonormal lists of eigenvectors of A,B, and A+B, respectively, such that Axi=λi(A)xi,Byi=λi(B)yi, and (A+B)zi=λi(A+B)zi for each i=1,…,n. For a given i∈{1,…,n} and any j∈{0,…,n−i}, let S1=span{x1,…,xi+j},S2=span{y1,…,yn−j}, and S3=span{zi,…,zn}. Then dimS1+dimS2+dimS3=(i+j)+(n−j)+(n−i+1)=2n+1 so (4.2.3) ensures that there is a unit vector x∈S1∩S2∩S3. Now invoke (4.2.2) three times to obtain the two inequalities λi(A+B)≤x∗(A+B)x=x∗Ax+x∗Bx≤λi+j(A)+λn−j(B) The first inequality follows from x∈S3 and the second inequality follows from x∈S1 and x∈S2, respectively. The statements about the cases of equality in (4.3.2a) follow from the cases of equality in (4.2.2) for the unit vector x and the inequalities x∗Ax≤ λi+j(A),x∈S1;x∗Bx≤λn−j(B),x∈S2; and λi(A+B)≤x∗(A+B)x,x∈S3. The inequalities (4.3.2b) and their cases of equality follow from applying (4.3.2a) to −A,−B, and −(A+B) and using (4.2.5): −λn−i+1(A+B)=λi(−A−B)≤λi+j(−A)+λn−j(−B)=−λn−i−j+1(A)−λj+1(B) 240 Hermitian matrices, symmetric matrices, and congruences If we set i′=n−i+1 and j′=j+1, the preceding inequality becomes λi′(A+B)≥λi′−j′+1(A)+λj′(B),j′=1,…,i′ which is (4.3.2b) If A and B have no common eigenvector, then the necessary conditions for equality in (4.3.2a,b) cannot be met. Weyl's theorem describes what can happen to the eigenvalues of a Hermitian matrix A if it is additively perturbed by a Hermitian matrix B. Various assumptions about the perturbing matrix B lead to inequalities that are special cases of (4.3.2a,b). In each of the following corollaries, we continue to use the same notation as in (4.3.1), and we continue to insist on the algebraic ordering (4.2.1) for all lists of eigenvalues. Exercise. Let B∈Mn be Hermitian. If B has exactly π positive eigenvalues and exactly v negative eigenvalues, explain why λn−π(B)≤0 and λv+1(B)≥0 with equality if and only if n>π+v, that is, if and only if B is singular.