Prove every isometry on an odd- dimensional real product space has 1 or -1 as an eigenvalue.

Question

This is a question from Axler. I was hoping for some help. It seems easy to understand, but I don't know where to go about on proving this.

Answer 1

Sure, let $d(v,w) = \displaystyle \sqrt{\langle v -w , v -w \rangle}$ be the metric on this real inner product space. Let $T$ be an isometry and assume that $Tv = \lambda v$ for some $v$ and $\lambda \in \mathbb{R}$. Consider $d(0,v) = \sqrt{\langle v, v \rangle}$. Since $T$ is an isometry, $d(0,v) = d(0, Tv) = \sqrt{\lambda^2 \langle v ,v \rangle }$ which forces $\lambda = \pm 1$.

You're using the odd dimension to ensure that $T$ has a real eigenvalue. This follows because the characteristic polynomial of $T$ will have real coefficients and odd degree and hence will have at least one real root.

Answer 2

$T$ is an isometry linear transformation, implies $||T(x) - T(y)||$ = $||x-y||$

i.e. $||T(x-y)||$ = $||x-y||$ for all $x$ , $y$ in V

or $||T(v)||=||v||$ for v in V

if v is an eigenvector then $||\lambda(v)|| = ||v||$

$|\lambda|$ $||v||= ||v||$

Now odd dimension says that there is one real eigenvalue so, $|\lambda|=1$ which gives $\lambda = 1 $ or $-1$.