Let $A$ where it's non-zero singular values are $\sigma_1,\ldots,\sigma_r$. $A$ can be written as $A=U\Sigma V$ where both $U,V$ are unitary matrices.
Let's look at $$AA^T = U\Sigma V^TV\Sigma U^T= U\Sigma^2 U^T$$
Now, from this point one should infer that the eigenvalues of $AA^T$ are $\sigma_1^2,\ldots,\sigma_r^2$ but I'm not sure how exactly. Is that has something to do with the fact that $U$ is unitary?
Best Answer
Suppose $A$ is an $m\times n$ matrix over $\mathbb{R}$ and consider the equation $AA^T = U\Sigma\Sigma^TU^T$, where $A = U\Sigma V^T$ is the SVD of $A$. Then $\Sigma\Sigma^T = \operatorname*{diag}\{\sigma_1^2,\ldots,\sigma_r^2,0,\ldots,0\} \in \mathbb{R}^{m\times m}$ where $r \leq \min\{m,n\}$ is the rank of $A$. Since $U$ is an orthgonal matrix we can write this equation equivalently as $$ (AA^T)U = U\Sigma\Sigma^T $$ Let $\lambda_i$ denote the $i$th diagonal element of $\Sigma\Sigma^T$. Then this equation says that $$ (AA^T)u_i = \lambda_i u_i $$ for $i = 1,\ldots,m$. Thus, each squared singular value $\sigma_j^2$ is indeed an eigenvalue of $AA^T$ with corresponding eigenvector $u_j$. The remaining eigenvalues of $AA^T$ (if there are any) is $0$ with algebraic multiplicity $m-r$.