Linear Algebra

Problem 1

Prove that if $T\in \mathcal{L}(V)$ is normal, then
$$\mbox{null}T^k=\mbox{null}T \mbox{ and } \mbox{range}T^k=\mbox{range}T$$

Solution:
easy to check that $\mbox{null}T\subset\mbox{null}T^k$
to prove an inclusion in the other direction, suppose that $v\in\mbox{null}T^k$
$T^kv=T(T^{k-1}v)=0\implies T^{k-1}v\in\mbox{null}T \mbox{ and } T^{k-1}v\in\mbox{range}T^{k-1}\subset\mbox{range}T$
but $V=\mbox{null}T\oplus(\mbox{null}T)^{\perp}=\mbox{null}T\oplus\mbox{range}T^*=\mbox{null}T\oplus\mbox{range}T$
the third equality holds because $T$ is normal
$T^{k-1}v\in \mbox{null}T\cap\mbox{range}T\implies T^{k-1}v=0$, inductively, we have $Tv=0$
This shows that $\mbox{null}T^{k}\subset\mbox{null}T$, completing the proof that $\mbox{null}T=\mbox{null}T^k$

Now we are going to show that $\mbox{range}T^k=\mbox{range}T$
$\mbox{range}T=\mbox{range}T^*=(\mbox{null}T)^{\perp}=(\mbox{null}T^k)^{\perp}=\mbox{range}(T^k)^*=\mbox{range}T^k$
first equality holds because T is normal, third equality is come from the previous proof, the last equality holds because $T^k$ is normal