What is the relationship between the singular values and the eigenvalues of $$AA^T$$ if $A$ is a square matrix (if that even matters)?
So what I did:
$$AA^T = USV^TVSU^T = USSU^T = US^2U^T$$
I read on google that the singular values of $A$ are the square root of the eigenvalues of $AA^T$ but I don't really know how to derive this from the equation above (or at all)?
Any hints how I can proof this?
Best Answer
That's the relationship exactly. If you take the eigen decomposition of $AA^T$ just rewrite it.
$$ AA^T = U \Sigma^{2} U^T = U \Lambda U^T $$
where $\Lambda$ is the matrix of eigenvalues for $AA^T$. If $\Sigma^{2} = \Lambda$ then
$$ \sigma_{i} = \sqrt{\lambda_{i}} $$
Note that $AA^T$ is positive semidefinite so the eigenvalues are non-negative. If you multiply a diagonal matrix with another diagonal matrix it just multiplies each element by the other one so this squares them.