# Prove that $\ker P(A^\ast)$ is an invariant subspace of $A$.

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.

• It suffices to note that if $A$ is normal, then $A^* = q(A)$ for some polynomial $q$. – Omnomnomnom Feb 14 at 19:48

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?