Is every normal operator on a finite-dimensional inner product space an orthogonal projection?

Question

I'm working through Axler's Linear Algebra Done Right and he asks for a proof that for a normal operator, $T$, on a finite inner product space, that $$\operatorname{null}T^k=\operatorname{null}T\quad\text{and}\quad \operatorname{range} T^k=\operatorname{range}T$$ for every positive integer $k$.

It seems to me that $\operatorname{range}T=(\operatorname{null}T)^\perp$ (which was shown for a normal operator) implies that $T$ is an orthogonal projection; and so the above is trivially true since $T^k=T$. Am I missing something? The few proofs that I've seen for the above seem to make it far more complicated and I can't find any statement that a normal operator is necessarily an orthogonal projection.

Answer 1

Thanks to David's counterexample, I see that I went wrong in my assumption that $range\,T=(null\,T)^\perp\implies (T$ is an orthogonal projection$)$; although the converse is true.

Answer 2

Lemma: $\DeclareMathOperator{\null}{null}$ $\DeclareMathOperator{\range}{range}$

Let $T : X \to X$ be a linear map. Then $$\null T \subseteq \null T^2 \subseteq \null T^3 \subseteq \ldots$$

Furthermore, if $\null T^m = \null T^{m+1}$, then

$$\null T^m = \null T^{m+1} = \null T^{m+2} = \null T^{m+3} = \ldots$$

Similarly $$\range T \supseteq \range T^2 \supseteq \range T^3 \supseteq \ldots$$

Furthermore, if $\range T^m = \range T^{m+1}$, then

$$\range T^m = \range T^{m+1} = \range T^{m+2} = \range T^{m+3} = \ldots$$

It suffices to prove that $\null T^2 = \null T$ since then by the lemma we have $\null T^k = \null T$ for all $k \in \mathbb{N}$, and by the rank-nullity theorem, $\range T^k = \range T$.

Let $x \in \null T^2$. If $T$ is normal then we have $X = \null T \oplus \range T$ so

$$x = u + v$$

where $u \in \null T$ and $v \in \range T$.

We have $v = x - u \in \null T^2$ so $0 = T^2v = T(Tv)$ implies that $Tv \in \null T$. Also $Tv \in \range T$ so $$Tv \in \null T \cap \range T = \{0\} \implies Tv = 0$$

Again, $Tv = 0$ implies $v \in \null T$ and also $v \in \range T$ by the assumption, therefore $v = 0$.

So $x = u \in \null T$. We conclude $\null T^2 \subseteq \null T$, and then using the lemma $\null T^2 = \null T$.