$A=S\circ O$ intuition and proof

Question

I'm reading the following Lemma:

Lemma: Every Invertible Matrix $A$ can be represented as a composition of a symmetric Matrix $S$ and an orthogonal matrix $O$, i.e. $A=S\circ O$.

What is the intuition behind this lemma and how can I prove it?

Answer 1

The intuition:

Any symmetric matrix can be decomposited into $Q^{-1}DQ$ where $Q$ is orthogonal (orthonormal, actually) and $D$ is diagonal. Then, $A = Q^{-1}DQO$.

Note that the product of two orthogonal matrices is also orthogonal. An orthogonal matrix can be regarded as a linear transformation that change the orthogonal axises of a plane into another orthogonal one without stretching it.

A diagonal matrix can be seen as a linear transformation that keep the directions of all axises but stretches (none or) some of them.

So, what's the intuition of $A = Q^{-1}DQO$?

Imagine a linear operator is some kind of a transformation of a plane, you can perform the transformation directly, or you can first transform your axises into another orthogonal one without stretching them, then, stretch them, and finally, transform into another orthogonal one

The proof

I will use the theory of singular value decomposition, it's an application of eigenvalue.

$A$ can be decomposited into $U\Sigma V^t$ where both $U$ and $V$ are orthogonal. To get the desired $S$, write $V^t$ as $U^tO$, or equvilently: $$V^tU = O$$ Note that for an orthogonal, $U^tU = UU^t = I$

Then, we have $$A = U\Sigma U^t O$$

One can check that $ U\Sigma U^t$ is symmetric.

Take $S = U\Sigma U^t$, completing the proof.