Question

Prove that a normal operator with real eigenvalues is self-adjoint. I know a matrix (operator) is normal if: N*N = NN*. Also, I know a matrix is self-adjoint (hermitian) if N = N*.

Is the proof as easy as: (N*N)* = N*N** = N*N=NN* (properties of adjoints and definition of normal).
(NN*)* = N**N* = NN* = N*N ?

It seems a little too easy. It doesn't seem like I'm proving N = N*

Answer 1

Wew! This is a spicy one.

Answer 2

I have never studied this subject in-depth but after some Google-Fu this looks useful:

https://www.math.ucdavis.edu/~anne/WQ2007/mat67-Ll-Spectral_Theorem.pdf

Good luck.

Answer 3

you just stated what a normal operator is

Answer 4

The normal operator T is self adjoint with real eigenvalues

Answer 5

Because
\[M(T) = \left[\begin{matrix}\lambda_1 &0&0&\cdots\\0&\lambda_2&0&\cdots\\\cdots\\\cdots &0&\cdots&\lambda_n\end{matrix}\right]\]

Answer 6

T is self adjoint iff T = T* but
\[(T)^* = \left[\begin{matrix}\bar\lambda_1 &0&0&\cdots\\0&\bar\lambda_2&0&\cdots\\\cdots\\\cdots &0&\cdots&\bar\lambda_n\end{matrix}\right]\]

Answer 7

In \(\mathbb R\), \(\lambda =\bar\lambda\)

Answer 8

So, T is self-adjoint.
Dat sit

Answer 9

Sorry, \(M(T) = M(T^*)\), lack of M in the front. :)

Answer 10

okay i think i havee a proof for u