if $T$ is normal then $\text{range} \ T = \text{range} \ T^*$

Question

Prove that if $T\in\mathcal{L}(V)$ is normal, that is $TT^*=T^*T,$ then: $$\text{range} \ T = \text{range} \ T^*$$

I know that if $T$ is normal then $||Tv||=||T^*v||$ which implies $\text{null} \ T = \text{null } \ T^*$, but couldn't get further.

Also, I tried to work out with $\text{range} \ T^*=(\text{null} \ T)^\bot$, but got stuck.

Any help appreciated.

Answer 1

Proof if $V$ is finite dimensional.

It is enough to show that $range(TT^*)=range(T)$. We have $range(TT^*)\subset range(T)$. $TT^*(x)=0$ implies that $\langle TT^*(x),x\rangle=\langle T^*T(x),x)\rangle=\langle T(x),T(x)\rangle=0$ implies that $T(x)=0$. We deduce that $ker(TT^*)\subset ker(T)$, we have $T(x)=0$ implies $T^*(T(x))=T(T^*(x))=0$, we deduce that $ker(T)\subset ker(TT^*)$ and henceforth $ker(T)=ker(T^*T)$. this implies that $dim(range(T))=dim(range(TT^*)$ since $range(TT^*)\subset range(T), range(TT^*)=range(T)$.

Answer 2

An alternative approach: you can prove easily that for the adjoint of a linear map $S\in\mathcal L(V,W)$ it holds that

$$\ker S=(\operatorname{range} S^*)^\bot,\quad \ker S^*=(\operatorname{range} S)^\bot,\quad (S^*)^*=S,\quad (U^\bot)^\bot=U\tag1$$

for some arbitrary subspace $U$. And we have the decomposition

$$V=\ker(T)\oplus \ker(T)^\bot=T(V)^\bot\oplus T(V)\tag2$$

that from $(1)$ becomes

$$V=\ker(T)\oplus T^*(V)=T(V)\oplus\ker(T^*)\tag3$$

Then is clear that $T(V)=TT^*(V)$ and $T^*(V)=T^*T(V)$, and because $T$ is normal we conclude that $T(V)=T^*(V)$.