Singular-value decomposition of matrix times its transpose

Question

Give an $m*n$ matrix $A$, I'm trying to find the SVD of $AA^T$ using the product of the summation form:

$ A = \sum_i{\sigma_iu_iv_i^T}$

I've gotten as far as multiplying A by

$ A^T = \sum_i{\sigma_iv_iu_i^T}$

Resulting in

$AA^T = \sum_i{\sigma_iu_iv_i^T}\sum_i{\sigma_iv_iu_i^T}$

Which can be re-written as:

$AA^T = \sum_j\sum_i{\sigma_i\sigma_ju_iv_i^Tv_ju_j^T}$

I'm not sure how to simplify from here and how to argue this is in-fact an SVD (what would be the sufficient conditions?)

Answer 1

The (right) singular vectors in SVD are orthonormal. Therefore, if $A$ is real, $v_i^Tv_j=\langle v_i,v_j\rangle$ is nonzero (and equal to $1$) if and only if $i=j$. It follows that $AA^T=\sum_i\sigma_i^2u_iu_i^T$ and this is a SVD for $AA^T$.