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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?

Hanno
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openspace
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1 Answers1

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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 $\,\langle A^*x|y\rangle = 0\:\forall x\in V.\,$

  1. What does this imply for $y$?
  2. What does it mean for $\,y\,$ if $\,0=\langle x|Ay\rangle\,$ holds for all $x\in V\,$?

Note that $\exists$ equivalent post Kernel of adjoint and orthogonal complement images .

Added in edit: Found 4 older posts, sorted by age:

  1. Geometric intuition of adjoint
  2. Is the formula $(\text{ker }A)^\perp=\text{im }A^T$ necessarily true?
  3. $\operatorname{Im} A = (\operatorname{ker} A^*)^\perp$
  4. image of adjoint equals orthogonal complement of kernel

Questions 2 through 4 have have got an accepted answer.

Hanno
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