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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?

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    What is the spectral decomposition of symmetric, positive semidefinite $\rm A A^{\top}$? – 2017-02-20
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    The columns of $U$ are orthonormal. So you can write $U=[\vec{u}_1 \: \vec{u}_2 ...]$. – 2017-02-20
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    If $U\Sigma^2U^T$ was an eigenvalue decomposition with eigenvalues you wrote, then what would be needed to hold for $U$? What does unitarity of $U$ give us? – 2017-02-20
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    Well, I've read here: http://www.cs.ubc.ca/~nickhar/W12/NotesMatrices.pdf that the spectral theorem says exactly that the values on the diagonal of $\Sigma$ are the corresponding eigenvalues for the column vectors of $U$. – 2017-02-20
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    @blueplusgreen have you seen "eigenvalue decompostion"? That is, have you seen equations of the form $$ A = PDP^{-1}? $$ – 2017-02-20

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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 $m-r$.