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:
a) 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 $g^{2}=f$.
b) If $f$ is bijective, then there exist an isometry $f_{1} \in \operatorname{End}_{F}(V)$ and a self-adjoint $f_{2} \in \operatorname{End}_{F}(V)$ with positive real eigenvalues, such that $f=f_{1} \circ f_{2}$.
I already solved a) but having trouble solving b). I guess I first have to proof that $f_1, f_2$ even exist and then proof $f=f_{1} \circ f_{2}$ but how do I do that? Maybe with help of a)?
Thanks for any help!
Let $f$ be an automorphism. $\tilde f\circ f$ is self adjoint. Thus $V$ has an orthonormal basis consisting of eigenvectors $v_1,...,v_n$. That means, $\tilde f\circ f$ is diagonalizable. In particular, all eigenvalues $\lambda_i$ are nonnegative real numbers, where $1\leq i\leq n$. But since $f$ is an automorphism, all eigenvalues are even nonzero. Now define a self-adjoint endomorphism $f_2:V\rightarrow V$ by $v_i\mapsto\sqrt{\lambda_i} v_i$. From the required property $f=f_1\circ f_2$ you can now construct $f_1$ by $f_1=f\circ f_2^{-1}$. The obtained endomorphism then actually forms an isometry.I leave the steps of the calculation to you.