Let $V$ be a complex inner product space with $\dim(V)<\infty$. Let $T$ be a normal operator ($TT^*=T^*T$). Show that $V$ is the direct sum of $\text{null}(T)$ and $\text{range}(T)$. Show that for any $S$, if $ST=TS$, then $ST^{*}=T^{*}S$.
For the first one, we know that the rank-nullity theorem
$$\dim(V) = \dim( \text{null}(T) ) + \dim( \text{range}(T) )$$
Is it enough to say that "$V$ is the direct sum of $\text{null}(T)$ and $\text{range}(T)$"?
Also, for the second one. I am not how to the first result to prove that. I found this is not a trivial result in functional analysis (called Fuglede's theorem.)
Fix the $T$, the result is certainly true for diagonalisable matrices. now the set of $V(ST-TS) = \{S\in \mathbf M_n(\mathbb C):ST-TS=0\}$ and $V(ST^*-T^*S) = \{S\in \mathbf M_n(\mathbb C):ST^*-T^*S=0\}$ are two closed complex analytic variety. From Schur's theorem, it is clear that the set of diagonalisable matrices is able matrices are dense in $\mathbf M_n(\mathbb C)$. Let $i$ and $i'$ be their closed topological immersion into $\mathbf M_n(\mathbb C)$, which agree on the intersection with $N$, which is still dense the two subspace respectively. Since immersion is continuous and the space is Hausdorff, the two immersions agree completely.