Consider the linear map $A:V\to V$ and its adjoint map $A^*$
(then $\langle Ax|y\rangle = \langle x|A^*y\rangle$ for any $x,y\in V$).
Now we want to prove that $\operatorname{Im} A^{*} = (\operatorname{Ker}A)^{\perp}$, that is, the image of the adjoint map is the orthogonal complement to the kernel of $A$. And the same about $\operatorname{Ker}A^* = (\operatorname{Im}A)^{\perp}$.
My attempt:
Let's say $(\operatorname{Ker}A)^{\perp} = \{x \in V |\langle x|u\rangle = 0\:\forall u\in\operatorname{Ker}A \}$.
Now $\operatorname{Im}A^* = \{y\in V | \exists x\in V : A^*x = y \}$. Now consider $\langle A^{*}x|y\rangle$ for all $x\in (\operatorname{Ker}A)^{\perp}, y\in\operatorname{Ker}A$. This dot product equals $\langle x|Ay\rangle = \langle x|0\rangle = 0$. So all $x$ are also mapped into $(\operatorname{Ker}A)^\perp$. So $\operatorname{Im}A^{*} = (\operatorname{Ker}A)^\perp$.
Is it correct?
Best Answer
I'd say you are close to it: When considering $\langle A^{*}x|y\rangle$ then $x$ should run through all of $V$, instead of being restricted to $(\operatorname{Ker}A)^\perp$.
A reformulation of yours: Consider some $y\in V$ such that $\,\langle A^*x|y\rangle = 0\:\forall x\in V.\,$
Note that $\exists$ equivalent post Kernel of adjoint and orthogonal complement images .
Added in edit: Found 4 older posts, sorted by age:
Questions 2 through 4 have have got an accepted answer.