Background
My undergrad education in linear algebra ended with eigenvalues, symmetric matrices, and orthogonal matrices. Now I'm trying to gain an intuitive understanding of singular values, and reached a new perspective/understanding of square matrices described below. However in case I have some misconceptions, or in case I am misguiding myself, I would like to test it.
Questions
Can every square matrix be expressed (uniquely?) as a product:
$$A=SR$$
where
- $S$ is a symmetric matrix (and therefore has orthogonal eigenspaces, like an inertia tensor or variance matrix, i.e. can be diagonalized by an orthogonal matrix)
- $R$ is an orthogonal matrix (and therefore is a distance-preserving rotation/reflection)
If yes, is this decomposition unique?
I like this perspective because:
- A symmetric matrix and an orthogonal matrix both exert actions associated with simple geometric intuitions.
- It makes evident what singular values are. (They are just the eigenvalues of $S$.)
- The geometric effect of the transpose of a matrix becomes more apparent: $A^T=R^{-1}S$. As does its relationship to the inverse $A^{-1}=R^{-1}S^{-1}$, as inverting a symmetric matrix corresponds to simple geometric intuitions (inverting each eigenvalue).
Any guidance/comment on a shortcoming, caveate, or misconception you see in my understanding would be appreciated, or any pointer to related concepts you think could be useful to be aware of, before I internalize this perspective.
Best Answer
Sure, use the SVD decomposition: if $A=U\Sigma V^\top$ for $U,V$ orthogonal and $\Sigma$ diagonal, then we can rearrange this to get $U\Sigma U^\top \cdot U V^\top$, while $U \Sigma U^\top$ is symmetric whose spectral decomposition is given, and $U V^\top$ is unitary as a product of such.
In general the geometric intuition of what you said is already evident from $U \Sigma V^\top$ without the need to do this trick, $A \vec{x}$ is just:
Projecting $\vec{x}$ onto the columns of $V$ to get $\vec{\pi}=V^\top \vec{x}$
We then stretch each entry $i$ of the result $\vec{\pi}$ by $\sigma_i$
Use the resulting vector entries as coordinates, to make a linear combination of the columns of $U$