Proving $\ker(T) = \ker(T^*)$ and $\operatorname{range}(T) = \operatorname{range}(T^*)$ for normal operator T

Question

Also, we have that $T:V \rightarrow V$ is a linear operator. Here, $T^*$ is the adjoint of $T$. I am unsure how to continue my proof. I've claimed that $\|T(v)\| = \|T^*(v)\|$ and have proven my claim, but not sure how to continue from there. Any help is appreciated, thanks

Answer 1

Since $\|Tv\| = \|T^*v\|, \forall v \in V$ we have $$v \in \ker T \iff Tv = 0 \iff \|Tv\| = 0 \iff \|T^*v\| = 0 \iff T^*v = 0 \iff v \in \ker T^*$$

so $\ker T = \ker T^*$.

Now we can also conclude

$$\operatorname{range} T = (\ker T^*)^\perp = (\ker T)^\perp = \operatorname{range} T^*$$