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.

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- Jan 14th 2011, 10:02 PMrishirichHilbert 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. - Jan 15th 2011, 12:29 AMOpalg
That is not correct. It should be: $\displaystyle T(M_1) \subseteq M_2$ if and only if $\displaystyle T^*(M_2^\perp) \subseteq M_1^\perp$.

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

. . . . . . . . . . .$\displaystyle \begin{aligned}T(M_1) \subseteq M_2\ &\Longleftrightarrow\ \langle Tx,y\rangle = 0 \quad(x\in M_1,\ y\in M_2^\perp) \\ &\Longleftrightarrow\ \langle x,T^*y\rangle = 0 \quad(x\in M_1,\ y\in M_2^\perp) \\ &\Longleftrightarrow\ T^*(M_2^\perp) \subseteq M_1^\perp .\end{aligned}$ - Jan 15th 2011, 10:04 PMrishirich
Hey thank you for your reply..Why is the question wrong?

- Jan 16th 2011, 12:53 AMOpalg
The simplest way to see that it is wrong is to think about what happens if $\displaystyle H_2=H_1$ and $\displaystyle M_2=M_1$. Then the question is saying that if a subspace is invariant under $\displaystyle T$ then it is also invariant under $\displaystyle T^*$. 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 $\displaystyle \begin{bmatrix}0&1\\0&0\end{bmatrix}$, but not under its adjoint.