Question

9. If 12122...21n are the eigenvalues of a Hermitian matrix A, show that XAx 12 = max q(x), An = min q(x) where g(x) = -12. Show that, furthermore, 1; = max 9(x) j= 2,3,...,n XEY; X+0 where Y, is the subspace of CM consisting of all vectors that are orthogonal to the eigenvectors corresponding to 11, ..., 1;-1.

Answer 1

Satura Solution: Since this geren matrin y Hermitian sol all real eigen values and these ratu can arranged in descending order. di 12 . ?da e value can be e enist Inn 70 Let & be an eigenvalue of n. Then there an eigen rector x such that Aa=xn. Tluua ŽAX = ilan) = Ton no] ... 9-(*) : **** So all the eigen value can be obzerined from qin) for suitable cheese of a. Thus the mine num value of q (n) is in and the manimum value of in) is ni. ie x = max qin) and in = min qlng no 2nd part: ut di = man ln let a gives the value of q. (r) *#0 je q (4) =21 Tiwe know that all the eigen vectors are orthogonal Let a and u are the eigenvalues of A and the Corresponding eigen reefer X, and up then Ах, = 2, Cowo 4, 2) X(X2 uy = {axy xx> = (An, "> = <", Atry> =< A>=<*y roguy =^<» ">

=) < Wp) = "<", "> =) (^-^) <" "p> =0 &t. than < "m> = o. 1 and up are ortrogonal to each other.] This another eigenvectoris ( corresponding to another eigenvalues) are of thogonal to x. Thurs another eigenvalues can be obtained from P(m) with the rectora x which are orthogonal. to x,. . Az = man q (n) .. . I antm 7 th x € Yz ( is the Subspace of cn of all rectors ortrogonal to the eigen vector Corresponding to del Similarly, t3 can be obtained from P(n), which takes maximum value for those to which are тоқта! tо ч аз х2 [Az - Pч (+) 7 In this way we get di a max g(n) where ncy Y; is the given subspace in questen. **0