$F \circ G$ self-adjoint $\iff F \circ G = G \circ F$

Question

Let $V$ be an finite inner product space and $F$ and $G$ two self-adjoint endomorphisms on $V$. Show that iff $F \circ G$ is self-adjoint, $F \circ G = G \circ F$.

My Proof:

"$\implies$": If $F \circ G$ is self-adjoint, we have $$ \langle F(G(v)), w \rangle = \langle v, F(G(w)) \rangle $$ and $$ \langle F(G(v)), w \rangle = \langle G(v), F(w) \rangle = \langle v, G(F(w)) \rangle $$ for all $v, w \in V$, because $F$ and $G$ are self-adjoint. To complete my proof I have to show

$$ \langle v, F(G(w)) \rangle = \langle v, G(F(w)) \rangle \implies F(G(w)) = G(F(w)),$$

but I don't know how.

The other direction didn't provide any obstacles for me.

Answer 1

Let $w$ an arbitrary given vector. The equality

$$\langle v, F(G(w)) \rangle = \langle v, G(F(w)) \rangle $$ for all vector $v$ is equivalent to

$$0= \langle v, F(G(w)) - G(F(w)) \rangle $$

but this means that $F(G(w)) - G(F(w))=0_V$ as it's the only vector orthogonal at all vector $v\in V$.

Remark: An alternative and straightforward way to prove the result is to use this

$$(F\circ G)^*=G^*\circ F^*=G\circ F$$

Answer 2

As $\langle v,F(G(w))\rangle=\langle v,G(F(w))\rangle$ for all $w,v\in V$, we have $\langle v,F(G(w))\rangle-\langle v,G(F(w))\rangle=0$ for all $w,v\in V$.

Now, let $v=F(G(w))-G(F(w))$. Then

$$\langle F(G(w))-G(F(w)),F(G(w))\rangle-\langle F(G(w))-G(F(w)),G(F(w))\rangle=\langle F(G(w))-G(F(w)),F(G(w))-G(F(w))\rangle=0$$

for all $w\in V$, by bilinearity(semi-bilinearity) of $\langle\cdot,\cdot\rangle$.

Thus by the positive definiteness of $\langle,\rangle$, i.e.

$$\langle v,v\rangle=0\text{ iff }v=\mathbf0 \text{ f.a. }v\in V$$

we have $F(G(w))-G(F(w))=\mathbf0$ f.a. $w\in V$, i.e. $F(G(w))=G(F(w))$ for all $w\in V$, i.e. $F\circ G=G\circ F$.