Let $V$ be a complex unitary vector space and $f$ an endomorphism on $V$. Show that following statements are equivalent.
- $f$ is self-adjoint and all eigenvalues of $f$ are are non-negative real numbers.
- There exists an endomorphism $g$ on $V$ with $f=g\circ \hat g$.
I think that in the $2\implies 1$ is rather simple to show $\hat f=\widehat{g\circ \hat g}=\hat{\hat g}\circ \hat g=g\circ \hat g$ from a previously proven statement but I'm having trouble proving the other way around. Can I simply just assume that such $g$ exists that satisfies the condition of $f$ being self-adjoint and then show that the composition is also self-adjoint?$$\langle f(v),w\rangle =\langle v,f(w)\rangle \implies \left \langle g\left (\hat g(v)\right ),w\right \rangle =\left \langle v,g\left (\hat g(w)\right )\right \rangle \implies \left ((g\circ \hat g )v\right )^\intercal \overline w=v^\intercal \overline{\left ((g\circ \hat g)w\right )}$$$$v^\intercal \left (g\circ \hat g\right )^\intercal \overline w=v^\intercal \overline{\left (g\circ \hat g\right )}\overline w.$$I'm not sure if what I've done so far is any good but this is about as far I got, I'd appreciate help.