- #1

CrazyIvan

- 45

- 0

## Homework Statement

Prove or give a counterexample: if [tex]\mathcal{S} \in \mathcal{L} \left( V \right) [/tex] and there exists and orthonormal basis [tex]\left( e_{1} , \ldots , e_{n} \right)[/tex] of [tex]V[/tex] such that [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex] for each [tex]e_{j}[/tex], then [tex]\mathcal{S}[/tex] is an isometry.

## Homework Equations

[tex] \mathcal{S} [/tex] is an isometry if

[tex]\left\| \mathcal{S} v \right\| = \left\| v \right\| [/tex]

for all [tex]v \in V[/tex].

## The Attempt at a Solution

I figure that if I can prove that [tex]\mathcal{S}[/tex] is normal, then I can use the Spectral Theorem to prove that all every basis vector is an eigenvector.

Then I can prove that the square of the absolute value of the eigenvalue for each eigenvector equals 1, and the proof comes easy after that.

But I'm getting hung up on proving if it is normal.

Thanks in advance for any help.