Let $(V,\langle\cdot,\cdot\rangle)$ a unitary vector space with $\dim(V)<\infty$ and $g\in L(V,V)$. Define $f:=g \circ g^\text{ad}\in L(V,V)$.
$i)$Show $f$ is self-adjoint.
$ii)$ all eigenvalues of f are non-negative real numbers.
$iii)$ Let $h\in L(V,V)$ self adjoint. Show: If all eigenvalues of h are non-negative, then there exists $k\in L(V,V)$ with $k=k^\text{ad}$ and $h=k\circ k$.
I was able to solve $i)$ and $ii), $ (my definition of scalar product is semi-linear in the first component.)
$$i): \langle f(v),w\rangle=\langle g\circ g^\text{ad}(v),w\rangle=\langle g^\text{ad}(v),g^\text{ad}(w)\rangle=\langle v,g^\text{ad}(g(w))\rangle=\langle v, f(w)\rangle $$
Let $\lambda$ a eigenvalue of $f$ with eigenvector $v$. I use $i)$ in the third equality $$ii):\overline \lambda \langle v,v\rangle=\langle \lambda v,v\rangle=\langle f(v),v\rangle= \langle v,f(v)\rangle=\langle v,\lambda v\rangle=\lambda\langle v,v\rangle$$ Thus $\lambda=\overline \lambda$ and therefore $\lambda$ is real. Suppose $\lambda$ is negative Then $$0>\lambda \langle v,v\rangle=\langle v,f(v)\rangle=\langle v,g(g^\text{ad}(v)\rangle=\langle g^\text{ad}(v),g^\text{ad}(v)\rangle>0$$ Contradiction, thus $\lambda$ is no negative.
I do not know how to start with $iii)$. Some approaches are welcome and a comment about $i)$ or $ii)$ is also welcome!