Lecture 16 - Finishing Isometries

We define an isometry:

isometry

An operator $S \in L (V)$ is called an isometry if:
$‖ S v ‖ = ‖ v ‖$
for all $v \in V$ .
In other words, an operator is an isometry if it preserves norms.

For instance, when is the operator $λ I$ an isometry. We need:

\begin{aligned} ⟨ λ I v, λ I v ⟩ & = ⟨ v, v ⟩ \forall v \in V \\ λ \overset{―}{λ} ⟨ v, v ⟩ & = ⟨ v, v ⟩ \forall v \in V \end{aligned}

Thus $λ \overset{―}{λ} = 1$ so then $| λ | = 1$ . So $λ$ lives in the unit circle on the complex plane.

But this was a specific operator. What about operators with more than that single $λ$ ? Can an operator with an orthonormal eigenbasis be an isometry? Suppose $S \in L (V)$ and we have an orthonormal eigenbasis via the (Complex/Real) Spectral Theorem; $e_{1}, . . ., e_{n}$ corresponding to the eigenvalue $λ_{1}, . . ., λ_{n}$ .

Note that:

v = \sum_{i = 1}^{n} ⟨ v, e_{i} ⟩ e_{i} \Rightarrow ‖ v ‖^{2} = \sum_{i = 1}^{n} | ⟨ v, e_{i} ⟩ |^{2}

Thus:

S v = \sum_{i = 1}^{n} λ_{i} ⟨ v, e_{i} ⟩ \Rightarrow ‖ S v ‖^{2} = \sum_{i = 1}^{n} | λ_{i} |^{2} | | ⟨ v, e_{i} ⟩ |^{2}

We want to require that $| v ‖ = ‖ S v ‖$ so then we desire that all $| λ_{i} |^{2} = 1$ so $| λ_{i} | = 1$ (note that theis has to be true for any $v$ chosen, so choosing $v = e_{1}, . . ., e_{n}$ each case yields the case above).

Characterization of isometries

$S$ is an isometry
$⟨ S u, S v ⟩ = ⟨ u, v ⟩$ for all $u, v \in V$
$S e_{1}, . . ., S e_{n}$ is orthonormal for every orthonormal list of vectors $e_{1}, . . ., e_{n}$ in $V$
$\exists$ an orthonormal basis $e_{1}, . . ., e_{n}$ of $V$ such that $S e_{1}, . . ., S e_{n}$ is orthonormal.
$S^{*} S = I$
$S S^{*} = I$
$S^{*}$ is an isometry
$S$ is invertible and $S^{- 1} = S^{*}$ .

Here think of:

(b) implies that $S$ preserves angles between $u, v$ since their angle output would be the same (as $S$ is an isometry so $‖ v ‖ = ‖ S v ‖$ and same for $u$ ).
The proof is in our chapter notes via:

Proof
We go (a) $\to$ (b) $\to$ $\dots$ $\to$ (h) $\to$ (a)