Recall:

In an essence, $S$ is a simple operator, so $T$ and $\sqrt{T^{\ast }T}$ are very similar. We want to look at the operator $\sqrt{T^{\ast }T}$ in more detail to work with, since we know a lot of its properties even when $T$ is a mess.

Notice that $\sqrt{T^{\ast }T}$ is a positive operator, so it must have non-negative eigenvalues. We used this to make:

In particular, there are exactly $\text{dim}(V)$ singular values, and they are all non-negative. Because of this, we can order them, so we typically write them in descending order:

or more traditionally:

An Example

Let $T\in \mathcal{L}({\mathbb{F}}^3)$ be defined by:

Thus:

Further:

Notice then:

Notice that we know that $T^{\ast }T$ is positive, as shown since $\mathcal{M}(T^{\ast }T,\beta )$ is symmetric. It has eigenvectors:

With eigenvalues:

So then the matrix with respect to the eigenbasis ${\beta }^{\prime }=\{e_1,e_2,e_3\}$ with each $e_i$ is the normalized $v_i$:

Thus:

Notice we have the singular values on the digagonal our singular values, namely:

So the singular value decomposition is as follows, but we'll need a definition.

Singular Value Decomposition

Notice it's doing a scaling and a swap. It swaps all the $e_i$'s with $f_i$'s and then scales respectively by $s_i$.

In general:

Notice that:

This implies that every operator has a matrix that is diagonalizable!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\begin{proof}
The operator $\sqrt{T^{\ast }T}$ is:

• positive
• self-adjoint

So there is an ONEB $e_1,...,e_n$ corresponding to the eigenvalues $s_1,...,s_n$ for $T$.

If our arbitrary $v=\sum _{i=1}^n⟨v,e_i⟩e_i$ then:

Apply the isometry $S$ (there may be many, but just pick one) from the polar decomposition to get:

where we can just define $f_i=S(e_i)$.
\end{proof}
In the matrix language, this is what Singular-Value Decomposition (SVD) looks like.

M = USV^

M = \sum\limits_{i=1}^{\min(m,n)} s_{i}\underbrace{\vec{u_{i}}\vec{v_i}^T}_{\text{an outer product}}

M_k = \sum\limits_{i=1}^{k} s_{i}\vec{u_{i}}\vec{v_i}^T