Normal Operators self adjoint

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This is one of my homework questions and I was hoping someone could check my proof:

Prove that a normal operator with real eigenvalues is self-adjoint.

So what I thought to do was, for some normal operator $N$, we can write $$N=UDU^*.$$ So \begin{align}N^* &= (UDU)^* \\ &=(U^*)^* D^* U^* \\ &= UD^* U^*\end{align} We can say that $D^*= \overline{D}$, and since this is real, $D^* = D.$ So $$N^*=UDU^*.$$ Is this correct?

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1 Answer

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Yes, it is correct, assuming you're only considering finite dimensional spaces.

However, you should specify at the outset that $D$ is supposed to be diagonal, which is possible because any normal matrix is unitarily diagonalizable.

You just have a typo in the line $$ N^*=(UDU)^* $$ which should be $$ N^*=(UDU^*)^* $$

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