1. ## Hilbert spaces

Let H1 and H2 be Hilbert spaces..If T:H1-->H2 is a bounded linear transformation with M1 closed subspace of H1 and M2 closed subspace of H2 then show that T(M1) is contained in M2 if and only if T*(M2) is contained in M1..

Any help appreciated.

2. Originally Posted by rishirich
Let $H_1$ and $H_2$ be Hilbert spaces. If $T:H_1\to H_2$ is a bounded linear transformation with $M_1$ closed subspace of $H_1$ and $M_2$ closed subspace of $H_2$ then show that $T(M_1) \subseteq M_2$ if and only if $T^*(M_2) \subseteq M_1$.
That is not correct. It should be: $T(M_1) \subseteq M_2$ if and only if $T^*(M_2^\perp) \subseteq M_1^\perp$.

The reason is that $M_2 = M_2^{\perp\perp}$ and so

. . . . . . . . . . . \begin{aligned}T(M_1) \subseteq M_2\ &\Longleftrightarrow\ \langle Tx,y\rangle = 0 \quad(x\in M_1,\ y\in M_2^\perp) \\ &\Longleftrightarrow\ \langle x,T^*y\rangle = 0 \quad(x\in M_1,\ y\in M_2^\perp) \\ &\Longleftrightarrow\ T^*(M_2^\perp) \subseteq M_1^\perp .\end{aligned}

The simplest way to see that it is wrong is to think about what happens if $H_2=H_1$ and $M_2=M_1$. Then the question is saying that if a subspace is invariant under $T$ then it is also invariant under $T^*$. That is certainly not true. For example, in a 2-dimensional space, the subspace spanned by the second basis vector is invariant under the transformation with matrix $\begin{bmatrix}0&1\\0&0\end{bmatrix}$, but not under its adjoint.