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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} \}$

Injectivity:

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.

Surjectivity:

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.

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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 $T'(x_2)=T(x_2)=0+x_2=T(x_1)+T(x_2)=Tx=y$.

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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, $T(x_1)=T(x)=y$.

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