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