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Let $\mathbb{V}$ a vector space with inner product and finite dimension.

Prove that, if $f \in \text{End}(\mathbb{V})$ then $\text{Im}(f) = \text{Im}(f \circ f^*)$, where $f^*$ is the adjoint operator of $f$

My attempt:

$\supseteq$) Let $v\in \text{Im}(f \circ f^*)$. Then, $\exists w \in \mathbb{V}$ such that $(f \circ f^*)(w) = v$. This means $f(f^*(w)) = v$ so we can take $\alpha = f^*(w)$ and then we have $f(\alpha) = v \Rightarrow v \in \text{Im}(f)$

$\subseteq$) Let $v \in \text{Im}(f)$. Then, $\exists w \in \mathbb{V}$ such that $f(w) = v$. We want to see $v\in \text{Im}(f \circ f^*)$ which happens iff $\exists w' \in \mathbb{V} \text{ such that } f(f^*(w')) = v$. In order to prove this, i would like to take $w'$ so that $f^*(w') = w$ but i can't find a way to accomplish this.

Any help to end this proof?

mechanodroid
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jscherman
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  • Note that you haven't used that it is $f^*$ anywhere, your first argument seems to go for any endomorphism – Andres Mejia Dec 18 '17 at 17:09
  • https://math.stackexchange.com/questions/349738/prove-rank-ata-rank-a-for-any-a-m-times-n – Andres Mejia Dec 18 '17 at 17:20
  • @AndresMejia I didn't get how the rank suffices to prove the images are actually the same. Or i don't see how to use it in order to complete the proof. Also, is the first part of the proof wrong? – jscherman Dec 18 '17 at 18:14
  • First part is correct. The answer shows what you are trying to prove – Andres Mejia Dec 18 '17 at 18:21

1 Answers1

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Denote the endomorphism as $A \in \operatorname{End}(V)$.

Instead of $\operatorname{Im}AA^* = \operatorname{Im}A$, let's first prove $\operatorname{Ker} AA^* = \operatorname{Ker} A^*$.

Assume $x \in \operatorname{Ker} A^*$.

Applying $A$ to both sides of $A^*x = 0$ gives $AA^*x = A(A^*x) = 0$. Hence $x \in \operatorname{Ker} AA^*$.

Conversely, assume $x \in \operatorname{Ker} AA^*$.

We have:

$$AA^*x = 0 \implies 0 = \langle AA^*x, x\rangle = \langle A^*x, A^*x\rangle = \|A^*x\|^2 \implies A^*x = 0$$

Hence $x \in \operatorname{Ker} A^*$.

Now, for any operator $A$ we have the decomposition $V = \operatorname{Im} A \oplus \operatorname{Ker} A^*$. Since $AA^*$ is self-adjoint, in particular we have:

$$\operatorname{Im} AA^* \oplus \operatorname{Ker} A^* = \operatorname{Im} AA^* \oplus \operatorname{Ker} AA^* = V = \operatorname{Im} A \oplus \operatorname{Ker} A^*$$

We conclude $\operatorname{Im} AA^* = (\operatorname{Ker} A^*)^\perp = \operatorname{Im} A$.

mechanodroid
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  • Why is that $\mathbb{V} = \text{Im }A ⊕ \text{Ker } A^*$? – jscherman Dec 18 '17 at 22:37
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    @jscherman Notice that: \begin{align} x \in \operatorname{Ker}A^* &\iff A^x = 0 \ &\iff \langle A^x, y\rangle = 0, \forall y \in V \ &\iff\langle x, Ay\rangle = 0, \forall y \in V \ &\iff x\perp \operatorname{Im}A\ &\iff x\in (\operatorname{Im}A)^\perp \end{align} Now just use that for any subspace $M \le V$ we have $V = M \oplus M^\perp$. – mechanodroid Dec 18 '17 at 22:44
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    @jscherman Actually, you can just formulate the argument like this: $$\operatorname{Im} AA^* = (\operatorname{Ker} AA^)^\perp = (\operatorname{Ker} A^)^\perp = \operatorname{Im} A $$ thus avoiding the decomposition. – mechanodroid Dec 18 '17 at 22:47