# Show the map between orthogonal complement and range is a bijection

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.

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