Closed subspaces and operators

Question

Let $\mathcal{H}$ be a Hilbert space and $\mathcal{H}_1$ be closed subspaces of $\mathcal{H}$. Suppose that $T\in B(\mathcal{H})$. Consider $\mathcal{H}_2:=T\mathcal{H}_1$. Is $T^*\mathcal{H}_2 \subset \mathcal{H}_1$?

Answer 1

In general is not true. For example you can consider $T$ self-adjoint operator and $H_1=\langle v \rangle$ where $v\neq 0$ is not an eighenvector of $T^2$ ($H_1$ is closed because every finite dimensional subspace of an Hilbert Space is closed).

In this way it is not possible that $T^*T H_1\subset H_1$ because if $T^2( \lambda v)$ is in $H_1$,with $\lambda\neq 0$, than there exists $\mu$ such that $T^2(\lambda v)=\mu v$ and so

$T^2(v)=\frac{\mu}{\lambda}v$

and it is a contraddiction.

Answer 2

The answer is no.

Consider $T : \mathbb{C}^2 \to \mathbb{C}^2$ given by the matrix $ \pmatrix{0 & 1 \\ 1 & 1}$. We have $T^* = T$.

Then $T (\operatorname{span}\{(1,0)\}) = \operatorname{span}\{(0,1)\}$, but $$T(\operatorname{span}\{(0,1)\}) = \operatorname{span}\{(1,1)\} \not\subseteq \operatorname{span}\{(1,0)\}$$