Kernel of powers of a normal operator I need to prove that if $A$ is a normal operator on a finite-dimensional inner product space then the kernel of $A^k$ coincides with the kernel of $A$.
I would say that one inclusion is always true and easy to prove, but I don't see how to prove the other.
 A: The proof is trivial if $A$ is non-singular, so we don't study this case.
First, for all matrices $ker A\subset \ker A^k$.
Second, for normal matrices it's quite  easy to show that $\ker A=\ker A^\ast$, $im A\bot \ker A$, and $im A^\ast\bot \ker A^\ast$ cf., for example, on wiki.
Since $im A^\ast\bot \ker A^\ast$  and $\dim (im A^\ast) + \dim (\ker A^\ast) = \dim V$ where $V$ is the vector space we are working in, we obtain that $\ker A^\ast = (im A^\ast)^\bot$, $(\ker A^\ast)^\bot =  im A^\ast$. Clearly, $\ker A  = (im A )^\bot$, $(\ker A )^\bot =  im A $.
In addition, if $A$ is normal, then so is $A^k$.
Let's show that $\ker A^2\subset \ker A$. Indeed, let's take $0\ne x\in \ker A^2$ and arbitrary $y$.
Then we can write $$0=(A^2x,y) = (Ax,A^\ast y),$$
i.e. $Ax\in (im A^\ast)^\bot.$ On the other hand, for any matrix $B$ the inclusion $ im B\subset (\ker B^\ast)^\bot$ holds. Thus,
$$Ax\in (im A^\ast)^\bot\cap (\ker A^\ast)^\bot=  im A^\ast  \cap  \ker A^\ast=0,$$ therefore $x\in \ker A$ and hence $\ker A^2= \ker A$. E
In the same spirit, let's suppose that $\ker A^k=\ker A$ and consider $x\in \ker A^{k+1}$. For all $y$ we can say
$$0=(A^kx,A^\ast y),$$
hence
$A^kx\in (im A^\ast)^\bot \cap  im A  = (ker A^\ast) \cap  im A  =  ker A   \cap im A =0$, so $x\in \ker A^k=\ker A$,
and we conclude that $\forall k>0\, \ker A^k=\ker A$.
A: The claim is also true in infinite dimensions:
Let $x\in\ker(A^2)$. Then for any $y\in H$ we have $\langle Ax,A^*y\rangle = \langle A^2x,y\rangle = 0$. Hence, $Ax\in (\operatorname{ran}A^*)^\perp$. But (using the normality of the operator)
$$
\overline{\operatorname{ran}A^*} = \overline{\operatorname{ran}(A^*A)} = \overline{\operatorname{ran}(AA^*)} = \overline{\operatorname{ran}A}.
$$
Thus, $Ax\in (\operatorname{ran}A)^\perp = \ker A^*$. Finally, $\|Ax\|^2 = \langle A^*Ax,x\rangle = 0$.
I guess the claim even holds if $A$ is unbounded, but I am not really familiar with unbounded normal operators.
