Lecture 18 - Polar Decompositions

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.

We talked about:

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 $S^{- 1} = S^{*}$ .

Note that property (b) shows that lengths are preserved over isometries, and thus angles are as well.

We proved a lot of these but not all of them via were proved.

We note that also since $S$ being an isometry implies $S^{*}$ is an isometry, then all the properties apply to $S^{*}$ consequently. But we'll prove something pretty interesting.

Lemma

$S$ is an isometry iff $S$ is normal (ONEB with every eigenvalue having $| λ | = 1$ ).

This implies that the ONEB is on the unit circle via the complex plane.

7.D: Polar Decomposition and Singular Value Decomposition

We defined this before and will need it:

square root

An operator $R$ is called a square root of an operator $T$ if $R^{2} = T$ .

Here we say that $\sqrt{T}$ is the positive square root of $T$ , and $R = \sqrt{T}$ . It denotes the unique positive operator. In other words, $\sqrt{T}$ is positive and $(\sqrt{T})^{2} = T$ .

polar decomposition

Given $T \in L (V)$ , then there exists an isometry $S \in L (V)$ such that:

T = S \sqrt{T^{*} T}

Note that we better have $T^{*} T$ be positive, as we are square rooting it. Consider the map $T^{*} T$ :

⟨ T^{*} T v, v ⟩ = ⟨ T v, T v ⟩ \geq 0

Thus $T^{*} T$ is positive. We are now required to find the $S$ that will make the above work.

Proof
The general idea is that:

Here, we are curious about when acting on $\sqrt{T^{*} T}$ we get vectors in it's $range (T^{*} T)$ , so what vectors from this set needs to be mapped to the correct spots. We'll build up an isometry $S$ . Suppose $S$ maps $\sqrt{T^{*} T} v$ to $T v$ .

We actually have a problem of being well defined here, because $\sqrt{T^{*} T}]$ may not be injective. It's possible to have:

\sqrt{T^{*} T} v_{1} = \sqrt{T^{*} T} v_{2}

while $v_{1} \neq v_{2}$ . But we are building specifically an isometry. So we have to ask:

Are norms preserved under the transformation $S$ ? Namely:

\begin{aligned} ‖ \sqrt{T^{*} T} v ‖ & = ‖ T v ‖ & ? \\ ‖ T v ‖^{2} & = ⟨ T v, T v ⟩ \\ = ⟨ T^{*} T v, v ⟩ \\ = ⟨ \sqrt{T^{*} T} \sqrt{T^{*} T} v, v ⟩ \\ = ⟨ \sqrt{T^{*} T} v, \sqrt{T^{*} T} v ⟩ & \sqrt{T^{*} T} is self-adjoint \\ = ‖ \sqrt{T^{*} T} v ‖^{2} \end{aligned}

So yes $S$ is norm-preserving.

Is $S$ well-defined?

To show it is. Suppose $\sqrt{T^{*} T} v_{1} = \sqrt{T^{*} T} v_{2}$ . We need to show that $T v_{1} = T v_{2}$ . Look at:

\begin{aligned} ‖ T v_{1} - T v_{2} ‖^{2} & = ‖ T (v_{1} - v_{2}) ‖^{2} \\ = ‖ \sqrt{T^{*} T} (v_{1} - v_{2}) ‖^{2} \\ = 0 \end{aligned}

So then $T v_{1} = T v_{2}$ .

The last thing to do is check $S$ is actually linear. But that's easy to check.

So at the moment, we have $S$ is a linear operator that preserves norms. Since it does this, then the only way to get the $\vec{0}$ as an output is to input a norm of 0, and thus must be the input zero-vector. Thus $S$ is injective. As a result, it's also surjective.

But what about vectors not starting from the range of $\sqrt{T^{*} T}$ ? Note that $dim (range (\sqrt{T^{*} T})) = dim (range (T))$ . Thus, starting with a basis for $range (\sqrt{T^{*} T})$ can be mapped to a relabeled basis for $range (T)$ . But we can extend this basis, say $e_{1}, . . ., e_{m}$ up to $e_{1}, . . ., e_{m}, . . ., e_{n}$ for each of the spaces to be a basis for $V$ .

The basis $e_{m + 1}, . . ., e_{n}$ is a basis for $range (\sqrt{T^{*} T})$ and $f_{m + 1}, . . ., f_{n}$ for $range (T)$ (these are a basis for the orthogonal complement of these spaces, by the way). Define $S e_{i} = f_{i}$ for these $i$ 's. We claim that this works still.

Is $S$ an isometry still? Notice:

\begin{aligned} ‖ S (\underset{any v \in V}{\underset{⏟}{\sqrt{T^{*} T} v + \sum_{i = m + 1}^{n} α_{i} e_{i}}}) ‖^{2} & = ‖ T v + \sum_{i = m + 1}^{n} α_{i} S e_{i} ‖^{2} \\ = ‖ T v + \sum_{i = m + 1}^{n} α_{i} f_{i} ‖^{2} \\ = ‖ T v ‖^{2} + \sum_{i = m + 1}^{n} | α_{i} |^{2} & Both vectors orthogonal, and basis is normal \\ = ‖ \sqrt{T^{*} T} v ‖^{2} + ‖ \sum_{i = m + 1}^{n} α_{i} e_{i} ‖^{2} \\ = ‖ \sqrt{T^{*} T} v + \sum_{i = m + 1}^{n} α_{i} e_{i} ‖^{2} \end{aligned}

Thus $S$ is an isometry.
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Singular Value Decomposition

Recall that $T = S \sqrt{T^{*} T}$ . As such, then because $S$ is so simple, let's look at the other operator $\sqrt{T^{*} T}$ instead.

singular values of $T$

The singular values of some $T \in L (V)$ are the eigenvalues of $\sqrt{T^{*} T}$ with each eigenvalue repeated $dim (E (λ, \sqrt{T^{*} T}))$ times.

Typically, since these singular values are all positive, then they are usually written in descending order:

S_{1} \geq \dots \geq S_{dim (V)}

σ_{1} \geq \dots \geq σ_{dim (V)}

The top way is Axler's way, while the $σ$ way is the most typically adopted format.