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Let $A$ be a normal linear operator on a finite dimensional unitary space and $P(x)$ a polynomial. Prove that $\ker P(A^{*})$ is an invariant subspace of $A$ (where $A^{*}$ is its adjoint operator). I tried using the identity $$ \ker A = \ker A^{*} $$ but I am stuck.

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    $\begingroup$ It suffices to note that if $A$ is normal, then $A^* = q(A)$ for some polynomial $q$. $\endgroup$ – Omnomnomnom Feb 14 at 19:48
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Without loss of generality, we can assume $A$ is a normal matrix and that the inner product is the standard one on $\mathbb{C}^n$. Then $A$ is diagonalizable with a unitary matrix: $A=UDU^*$, so $A^*=UD^*U^*$.

Therefore $P(A^*)=UP(D^*)U^*$. A vector $v\in\ker P(A^*)$ if and only if $P(D^*)U^*v=0$. Saying that $\ker P(A^*)$ is $A$-invariant means that for such $v$, $Av\in\ker P(A^*)$, that is, $$ 0=P(D^*)Av=P(D^*)U^*UDU^*v=P(D^*)DU^*v $$ Thus we are reduced to prove that $\ker P(D^*)$ is $D$-invariant. Can you finish?

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