Let $F=\mathbb{R}$ or $F=\mathbb{C}$, let $(V,\langle.,.\rangle)$ be a finitely generated $K$-vector space with scalar product, let $f \in \operatorname{End}_{F}(V)$. Show that:
If $f$ is self-adjoint and all eigenvalues of $f$ are non-negative, then there exists a self-adjoint $g \in \operatorname{End}_{F}(V)$ with $g^{2}=f$.
Attempt:
Let $v \in V$ and $\lambda \in \mathbb{F}$. Since $f$ is self-adjoint, there exists an orthonormal basis $v_1,...,v_n$ of $V$ consisting of eigenvectors for $f$. Since all eigenvalues are non-negative, we have
$(*)$ $g(v)= \sqrt{\lambda}\cdot v$ and thus $(g \circ g)(v) = g(g(v)) = g(\sqrt{\lambda}\cdot v) = \lambda \cdot v = f(v)$.
Problem:
I got the feeling that I don't use that $f$ is self-adjoint in $(*)$. So this even necessary?
Thanks for any help!