If $V$ is an inner product space , $T:V\to V$ linear transformation, show that $(\ker(T))^{\bot}=\operatorname{Im}(T^*).$

Question

If $V$ is an inner product space, $T:V\to V$ linear transformation, show that

$(\ker(T))^{\bot}=\operatorname{Im}(T^*)$

when $T^*=\overline{(T)^T}. $

I know that each vector $x\in V$can be written as $k+t=x$ when $k\in (\ker(T))^{\bot}$ and $t\in \ker(T)$ - can it help somehow?

I tried to show in one direction $ \operatorname{Im}(T^*)\subseteq (\ker(T))^{\bot}$ but got stuck. How do I continue?

in othe direction $(\ker(T))^{\bot}\subseteq \operatorname{Im}(T^*) $ :

lets show that if $v\in (\ker(T))^{\bot}$ that exists $x\in V$ so that $T^*x=v$?

I take $u\in (\ker(T)),x\in V$

so $0=\langle x,0\rangle =\langle x,Tu\rangle =\langle T^*x,u\rangle$ so $T^*x \in (\ker(T))^{\bot}$ like I wanted

what do you think?

Answer 1

The key property you need here is:

$$\forall x,y \in V,\; \langle T^*(x), y\rangle = \langle x, T(y)\rangle$$

Thus, if $z\in \text{Im}(T^*)$, then $\exists x\in V,\;z = T^*(x)$ and you can see that $z\in (\text{Ker}(T))^{\perp}$ since $\forall y\in\text{Ker}(T)$: $$\langle T^*(x), y\rangle = \langle x, T(y)\rangle = \langle x,0\rangle = 0.$$

Hence, $\text{Im}(T^*)\subseteq (\text{Ker}(T))^{\perp}$.

Answer 2

First let me say that $\ker(T)^{\bot}=\operatorname{Im}(T^*)$ is not true in general for infinite dimensional $V$. For example the operator $T:(x_n)\mapsto(\frac {x_n}{2^n} )$ on $l_2$ is selfadjoint, injective but not surjective.

What is true in general is $\ker(T)=\operatorname{Im}(T^*)^{\bot}$:

$$x\in\ker T\iff T(x)=0\iff\forall y\in V:\space\langle T(x),y\rangle=\langle x,T^*(y)\rangle=0\iff x\in \operatorname{Im}(T^*)^{\bot} $$

Now if $V$ is $\textbf{finite dimensional}$ then $B^{{\bot}{\bot}}=B $ for any subspace $B\subseteq V$ so the above equality is equivalent to $\ker(T)^{\bot}=\operatorname{Im}(T^*)$.