This question is motivated by the following proposition from the book "Connections, curvature and cohomology" by Werner Greub:
If $(\xi, g)$, $(\eta, h)$ are riemannian vector bundles over the same base $B$ and $\varphi \in \text{Hom}(\xi, \eta)$ is an isomorphism, then there exists an isometric isomorphism $\psi : \xi \to \eta$.
If anyone has a proof of this fact, I will accept it. Otherwise, let me state the problematic point I had into simple linear algebra language.
Let $V$ be a finite dimensional real vector space and consider $g, h : V \times V \to \mathbb{R}$ two inner products. Let $A : V \to V$ a linear operator such that $$h(A(v_1), v_2) = g(v_1, v_2), \quad \forall v_1, v_2, \in V. $$
It is easily seen that $A : (V, h) \to (V, h)$ is self-adjoint and positive. Therefore, there exists a square root of $A$, say $B : V \to V$, with $B^2 = A$.
In this setting, how does one prove that $B : (V, g) \to (V, h)$ is an isometry?
I appreciate your help.
I solved it. It was simpler than I expected.
The square root of a positive operator is positive as well. In particular, $B$ is positive, hence self-adjoint (with respect to $h$). Then $$h(B(u), B(v)) = h(B^*B(u), v) = h(B^2(u), v) = h(A(u), v) = g(u,v), \quad \forall u, v \in V$$ as desired.