Let $T$ be a bounded linear operator on a Hilbert space. If $T$ is normal, how do I prove that null space of $T$ equals the null space of $T^2$?

Question

I can prove that $N(T)=N(T^*).$ How do I proceed from there?

Answer 1

What you get from $N(T)=N(T^*)$ is that $$ R(T)=N(T^*)^\perp=N(T)^\perp. $$ So, if $T^2x=0$, you have that $Tx\in N(T)\cap R(T)=\{0\}$. Thus $Tx=0$.

Answer 2

From what you know \begin{align} T^2x & = 0 \iff T^*Tx = 0 \\ & \implies \langle T^*Tx,x\rangle = 0 \\ & \implies \langle Tx,Tx\rangle =0 \\ & \implies Tx = 0 \end{align} The opposite implication that $Tx=0 \implies T^2x=0$ is clear.