Let $H,K$ be Hilbert spaces.

Suppose $T\in B(H,K)$.

Show that the restriction map of $T$, $T': \mathcal{N}^{\perp}(T) \rightarrow \mathcal{R}(T) $ is a bijection.

My attempt:

Note that $ \mathcal{N}^{\perp}(T) = \{ h\in H: \langle h, k \rangle=0, \forall k\in \mathcal{N} \}$


Suppose $y_1=y_2 \in \mathcal{R}(T)$. Then there exists $x_1,x_2 \in H$ such that $Tx_1=Tx_2$.

But what we want to show is that $x_1,x_2 \in \mathcal{N}^{\perp}(T)$.

So we want show that $\langle x_1,k \rangle = \langle x_2,k \rangle=0 $ for all $k$.

And this is where I got stuck.


Let $y \in \mathcal{R}(T)$. So there exists $x\in H$ such that $Tx=y$.

But we want to show that $x\in \mathcal{N}^{\perp}(T)$.

Again I'm stuck at this point.

Any help will be appreciated.


Injectivity: Suppose $x,y \in \mathcal N^{\perp} (T)$ and $T'x=T'y$. Then $Tx=Ty$ so $x-y \in \mathcal N (T)$. Hence, $ \langle (x-y), (x-y) \rangle =\langle x, (x-y) \rangle-\langle y, (x-y) \rangle=0-0=0$. This implies that $\|x-y\|=0$ so $x=y$,

Surjectivity: Let $y \in \mathcal R (T)$. Then there exists $x$ such that $y=Tx$. By a standard result in Hilbert space theory we can write $x$ as $x_1+x_2$ where $x_1\in \mathcal N(T)$ and $x_2 \in \mathcal N^{\perp} (T)$. Now $T'(x_2)=T(x_2)=0+x_2=T(x_1)+T(x_2)=Tx=y$.


Injectivity: suppose $T(x_1)=T(x_2)$, with $x_1,x_2\in\mathcal{N}^\perp$. Since $T(x_1-x_2)=T(x_1)-T(x_2)=0$, $x_1-x_2\in\mathcal{N}$, Therefore $x_1-x_2\in\mathcal{N}\cap\mathcal{N}^\perp$ and hence $x_1-x_2=0$.

Surjectivity. Take $y\in\mathcal{R}(T)$. Then $y=T(x)$ for some $x\in H$. Decompose $x=x_1+z_1$ where $z_1\in\mathcal{N}$ and $x_1\in\mathcal{N}^\perp$.Clearly, $T(x_1)=T(x)=y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.