How to prove that $\text{Ker}(A)=\text{Ker}(A^2)$ if $A$ is a normal operator?

Question

Prove that $\text{Ker}(A)=\text{Ker}(A^2)$ if $A$ is a normal operator.

I know that that $\text{Ker}(A)\subseteq \text{Ker}(A^2)$, which I proved as follows: $x\in \text{Ker}(A) \implies Ax=0\implies(AA)x=A(Ax)=0\implies x\in \text{Ker}(A^2)$

Answer 1

While this should be clear from the spectrum theorem, it can be established directly.

If $A^2x=0$, then $$0=(A^2x, A^2x)=(Ax, A^*A^2x)=(Ax, AA^*Ax)=(A^*Ax, A^*Ax)$$

Thus $A^*Ax=0$, hence $$0=(A^*Ax, x)=(Ax, Ax)$$

So $Ax=0$.

Answer 2

Any normal operator $A$ has the same kernel as its adjoint (since $\|Ax\|^2=(A^*Ax,x)=(AA^*x,x)=\|A^*x\|^2$), hence

$$A^2x=0\Rightarrow Ax\in\ker A=\ker A^*\Rightarrow0=(0,x)=(A^*Ax,x)=\|Ax\|^2.$$