Prove $f$ is a isometry

Question

Let $V$ be a real vector space with dimension $n$ and let $\beta$ be a positive bilinear form. Let $f$ be a endomorphism of $V$ which transforms orthogonal vectors into orthogonal vectors. Show that exists an isometry $g$ and a number $\lambda$ such that $f = \lambda g$

I tryied starting saying that $g$ and $\lambda$ exists only if $f$ is a isometry, so

$$\beta(v, w) = \beta(f(v), f(w))$$ but I'm not sure this is the best way to solve this problem.

Is there any better way?

Answer 1

Hint: Let $f^*$ denote the adjoint of $f$ relative to $\beta$. The following are equivalent:

1. $f$ transforms orthogonal vectors to orthogonal vectors,
2. For $v,w \in V$: $\beta(v,w) = 0 \implies \beta(f(v),f(w)) = 0$,
3. For $v,w \in V$: $\beta(v,w) = 0 \implies \beta(v,f^*(f(w))) = 0$.

Using statement 3 above, show that $f^* \circ f$ must be a multiple of the identity operator. Why does this imply that $f$ is a multiple of an isometry?

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You have now established that $f^*\circ f$ is a multiple of the identity. That is, there exists a $c$ for which $f^*(f(x)) = cx$. Note that we must have $c \geq 0$ because for some non-zero $v \in V$, we have $$ c = \frac{\beta(v,f^*f(v))}{\beta(v,v)} = \frac{\beta(f(v),f(v))}{\beta(v,v)} \geq 0. $$ With that established, let $g = \frac 1{\sqrt{c}} f$; we see that $g^* \circ g$ is the identity map. It follows that for all $v$ and $w$, we have $$ \beta(g(v),g(w)) = \beta(v,g^*(g(w))) = \beta(v,w), $$ so that $g$ is indeed an isometry.