Let $B, T\in \mathcal{B}(H)$ such that $NT=TN$, with $N$ is a normal and compact operator. Prove that all autospace of $N$ is reducing for $T$.

Question

One closed subspace $W\subset H$ is reducing for an operator $T:H\rightarrow H$, if $T(W)\subseteq W$ and $T(W^{\perp})\subseteq W^{\perp}$.

That's it, we want show that all autospace of $N$, $Aut_{N}(\lambda)=\{x\in H: N(x)=\lambda x\}$ is reducing for $T$ where $\lambda$ is a eigenvalue of $N$, i. e., we want show that $$T(Aut_{N}(\lambda))\subseteq Aut_{N}(\lambda)\quad and \quad T(Aut_{N}(\lambda)^{\perp})\subseteq Aut_{N}(\lambda)^{\perp},$$ for all $\lambda$ eigenvalue of $N$.

The first continence is possible through the commute for $N$ and $T$. Because if $x\in Aut_{N}(\lambda),$ then $$N(T(x))=NT(x)=TN(x)=T(\lambda x)=\lambda T(x), $$ what imply in $T(x)\in Aut_{N}(\lambda)$. And i know too that $TN^{*}=N^{*}T$, where $N^{*}$ is a adjoint of operator $N$. But, I don't know what to do to show $T(Aut_{N}(\lambda)^{\perp})\subseteq Aut_{N}(\lambda)^{\perp}$...

Answer 1

I will build further on your attempt and I'll show $T(\operatorname{Aut}_N(\lambda)^\perp) \subseteq \operatorname{Aut}_N(\lambda)^\perp$ provided that $TN^* =N^*T$, i.e. $NT^* = T^*N$.

Let $x \in \operatorname{Aut}_N(\lambda)^\perp$. Then if $y \in \operatorname{Aut}_N(\lambda)$ we must show that $\langle Tx, y\rangle = 0$.

However, note that $T^*y \in \operatorname{Aut}_N(\lambda)$, since $$NT^*y = T^*N y = T^* \lambda y = \lambda T^*y$$

Hence, $$\langle Tx,y\rangle = \langle x,T^*y\rangle = 0$$ since $x \in \operatorname{Aut}_N(\lambda)^\perp$ and $T^*y \in \operatorname{Aut}_N(\lambda)$.