I have a question regarding reducing subspaces:
Let $H$ be a complex Hilbertspace and let $T$ be a bounded linear operator on $H$. Furthermore, let $H_1$ be a $T$-reducing closed subspace, which means that $T(H_1)\subset H_1$ as well as $T(H_{1}^{\perp})\subset H_{1}^{\perp}$.
The statement I wanted to prove is: If $M$ is a closed $T\restriction_{H_1}$-reducing subspace of $H_1$, then $M$ is a $T$-reducing subspace.
Here's my approach: We denote the orthogonal complement with respect to a subspace $N$ by $^{\perp_{N}}$. For $M\subset H_1$ closed and $T\restriction_{H_1}$-reducing it follows that
$T(M)=T\restriction_{H_1}(M)\subset M$
as well as
$T(M^{\perp_{H}})=T((M^{\perp_{H}}\cap H_1) + (M^{\perp_{H}}\cap H_{1}^{\perp_{H}}))=T(M^{\perp_{H}}\cap H_1)+T(M^{\perp_{H}}\cap H_{1}^{\perp_{H}})=T(M^{\perp_{H}}\cap H_1)+T(H_{1}^{\perp_H})\subset T\restriction_{H_1}(M^{\perp_{H_1}})+H_{1}^{\perp_H}\subset M^{\perp_{H_1}} + M^{\perp_H}=(M^{\perp_{H}}\cap H_1)+M^{\perp_H}\subset M^{\perp_{H}}$
Does this work out? If so, I was wondering if this is the easiest way to show it, because it seems to me there's computational effort I could reduce, but I'm not certain. thanks in advance
Your proof is fine. I prefer the following. By assumptions $\mathcal{H}=H_1\oplus H_2$ and $H_1=M\oplus N,$ where $\oplus$ denotes the orthogonal sum, and all the spaces are closed and invariant for the operator $T.$ Then $\mathcal{H}=M\oplus (N\oplus H_2),$ and both direct summands are invariant for $T.$