Show that $|\lambda|\leq 1$ for each eigenvalue $\lambda$ of a partial isometry

Question

Let $V$ be an inner product space. A linear map $U:V\rightarrow V$ is a partial isometry if there is a subspace $M\subset V$ such that $\parallel Ux\parallel =\parallel x \parallel$ for all $x\in M$ and $\parallel Ux\parallel =0$ for all $x\in M^{\perp}$.

Prove that if $\lambda$ is a eigenvalue of a partial isometry then $|\lambda|\leq 1$.

This problem was taken from Halmos's Finite Dimensional Vector Spaces (sec. 76).

Thanks.

Answer 1

Hint: what can you say about $\|U\|$? That's usually the first thing one should look when trying to estimate a spectral radius.

Recall $V=M\oplus M^\perp$ for any subspace $M$ of a finite-dimensional inner-product space (or closed subspace of a Hilbert space, if you care about generalizing this fact). Then for every $v\in V$, we can write $v=x+y$ with $x\in M$ and $y\in M^\perp$. Hence $$\|v\|^2=\|x\|^2+\|y||^2\qquad Uv=Ux+Uy=Ux\quad\Rightarrow\quad \|Uv\|=\|Ux\|=\|x\|\leq \|v||.$$ Now just apply this to any eigenpair $(\lambda,v)$.