Question

Prove that a normal operator on a complex inner product space is self- adjoint if and only if all its eigenvalues are real. [The exercise above strengthens the analogy (for normal operators) between self-adjoint operators and real numbers.]

Answer 1

- self adjoint operator. An Operator TE ICV) is called self-adjoint es q = T* . 'In other words, Tedcr) is : self adjoint if and only it CTV, w) = Cv, TW) For all viwfV.

To prove that a normal operator on a complex inner product space is sells adjoint if and only in all its eigenvalles are real. Proof: 1. Let e a normal operator on a complex inner product space ie Te LCV) Seppose TEL (V) is sely adjoint operator . (Tv,w) = CV.TW) i for all viWEV. jc To I* let decis an eigenvalele of T and let Veo, Ver the corresponding eigen vector such that . To = tv

Since IV - TV dll V112 ET 117112 - A<vive a Taviry. <av, v kt Viv> <vidvs. <V, TV <vidv} = < V, TV) --- Csince T* T) ī <vers : t <viry I 11v112 = TKIVIR - ② from equation © and ☺ Aline = T1N112 da - Hence is real . e for all eigen values f or T are real. Now conversla assume that alu apgenvalues of T are real. To prove T is sent adjoint. . Let T is o normal operator on a complex. inner product space v then assume T is diagonalizable. T There is a basis of orthogonal eigenveetors vi and in that basis I is represented by a diagonal matrix with the eigenvalues a on the diagonal . The adjoint of T is represented in the same basis by a diagonal matrix with entries ti - But all dj are real _ dj = di . TET* Hence T is self adjoint.