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

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?

• Is $T^T$ the transpose of $T$? – Ruben Jun 24 at 16:17
• I believe that is correct. – paulinho Jun 24 at 16:18

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

• Thanks ! And how should I do the other direction? – KIMKES1232 Jun 24 at 16:35
• I added a suggestion to other direction what do u think ? – KIMKES1232 Jun 24 at 16:57
• Your suggestion is not correct. You should begin with $u\in (\text{Ker}(T))^{\perp}$, not in $\text{Ker}(T)$. – paf Jun 24 at 19:42

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