Two normal operator that commutes Suppose $N\in B(H)$ is normal, and $T\in B(H)$ is invertible. Prove that if $TNT^{-1}$ is normal then $N$ commutes with $T^*T$.
I can not any idea to prove it, just I know  $T^*TNT^{-1}{T^*}^{-1}N^*T^*T =N^*T^*TN$, because $TNT^{-1}$ is normal. 
Please just give me a hint .
Thanks in advance.
 A: Because $V = TNT^{-1}$ is normal and $N$ is normal, then
$$
           V = \int \lambda dE_{V}(\lambda),\;\;\; N=\int\lambda dE_{N}(\lambda),
$$
and $E_{V}(S) = TE_{N}(S)T^{-1}$ for all Borel subsets $S$. Hence,
$$
  TE_{N}(S)T^{-1} = E_{V}(S)=E_{V}(S)^{\star}=(T^{-1})^{\star}E_{N}(S)T^{\star}.
$$
Therefore,
$$
                 T^{\star}TE_{N}=E_{N}T^{\star}T.
$$
There are details for you to fill in, as requested.
A: We have $N^*=N^{-1}$ and ${T^*}^{-1}N^*T^*=TN^*T^{-1}$. Multiply by $T^{-1}$ to the left and ${T^*}^{-1}$ to the right. One gets
$$T^{-1}{T^*}^{-1}N^*=N^*T^{-1}{T^*}^{-1}$$
Bearing in mind that $T^{-1}{T^*}^{-1}=\left(T^*T\right)^{-1}$, one has
$$\left(T^*T\right)^{-1}N^{-1}=N^{-1}\left(T^*T\right)^{-1}$$
And we're done because $A^{-1}B^{-1}=B^{-1}A^{-1}\Rightarrow AB=BA$
A: Here is an interesting observation:
if $v \in \ker(N)$ then $T^*T v \in \ker(N)$
Proof:  keep in mind that $\ker(N)=\ker(N^*)$ and that this holds for all normal operators. Let $v \in \ker(N)$ be any vector. Since $T$ is invertible there is $\tilde{v}$ such that $v = T^{-1}\tilde{v}$. Then $T N T^{-1} \tilde{v} = 0$. But being $T N T^{-1}$ normal we get $(T N T^{-1})^* \tilde{v} = 0$ hence $$(T^*)^{-1} N^* T^*\tilde{v} = 0$$ so $$T^*\tilde{v} \in \ker(N)$$ that is $T^*Tv \in \ker(N)$ as we wanted to show. 
This solves the problem for finite dimensional spaces. Indeed, for any $\lambda \in \mathbb{C}$ consider the normal operator $N_{\lambda} := N - \lambda Id$. Then $T N_{\lambda} T^{-1}$ is normal hence (by the above applied to $N_{\lambda}$) we get $T^*T$ preserves $\ker(N_{\lambda})$ which are (for the adequated $\lambda$'s) the eigenspaces of $N$. This implies that $N$ commutes with $T^*T$. Indeed, the vectors space $H$ is a finite sum of eigenspaces $V_{\lambda_j}:= \ker(N_{\lambda_j})$, $j=1,\cdots,k$. i.e.
$$H = \sum_{j=1}^k V_{\lambda_j} \,.$$
Let $v \in H$ arbitrary and $v = \sum_{j=1}^k v_j$ its decomposition according to the above decomposition of $H$. Then $$ \begin{aligned} T^*T N v &= \sum_{j=1}^k T^*T N v_j = \sum_{j=1}^k T^*T \lambda_j v_j\\ &= \sum_{j=1}^k \lambda_j T^*T v_j = \sum_{j=1}^k N T^*T v_j \\ &= N T^*T \sum_{j=1}^k v_j = N T^*T v \end{aligned}$$
since $v \in H$ is arbitrary we get $T^*T N = N T^*T $  .
