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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.

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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 $(\lambda,v)$.

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  • $\begingroup$ I was going to post the same answer. Good hint. $\endgroup$
    – Mhenni Benghorbal
    Jul 5, 2013 at 0:31
  • $\begingroup$ @MhenniBenghorbal Thank you, Mhenni. $\endgroup$
    – Julien
    Jul 5, 2013 at 0:35
  • $\begingroup$ Minor technicality: In the infinite dimensional case it should be assumed that $M$ is closed (or else we can just use its closure). Although, given the title of the book, perhaps $V$ is implicitly assumed to be finite dimensional. $\endgroup$
    – Jonas Meyer
    Jul 5, 2013 at 0:40
  • $\begingroup$ @JonasMeyer Right, but it is taken from Halmos FDVS. So that's why I decided to let this go. Thanks for the note. $\endgroup$
    – Julien
    Jul 5, 2013 at 0:42
  • $\begingroup$ What happens if $V$ is infinite dimensional and is not closed? $\endgroup$
    – Pedro
    Jul 5, 2013 at 1:02

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