Realation between SVD and EVD Given SVD decomposition $ A=V\Sigma U^T $ (U and V are Orthonormal and $ \Sigma $ is a diagonal matrix), i wish to prove that $ AA^T=U\Sigma \Sigma ^TU^T $ is EVD decomposition of $ AA^T  $ (same goes for $ A^TA=V^T \Sigma ^T \Sigma V $).
Its easy to see that indeed $ AA^T=U\Sigma \Sigma ^TU $. But i dont understand why the values on  $ \Sigma \Sigma ^T $'s diagonal are $ AA^T $'s eigenvalues.
 A: You forgot a transpose: $A A^T = U \Sigma \Sigma^T U^T$. With that said, recall that by assumption $U^T = U^{-1}$. Now you can see by direct substitution that the columns of $U$ are the eigenvectors of $A A^T$ and that the eigenvalues are the diagonal entries of $\Sigma \Sigma^T$. From a more algebraic point of view, if you can similarity-transform a (square) matrix into diagonal form, then the diagonal entries of that diagonal matrix must be its eigenvalues.
The situation is slightly different for the "economy" SVD, but still essentially the same.
A: We can clearly show that by matrix transformations:
$$A = U\Sigma V^{T}$$
Using $(AB)^T = B^TA^T$:
$$AA^T = U\Sigma V^TV\Sigma ^TU^T$$
Using $V^TV = I$ for unitary matrix $V$, and $\Sigma $ being diagonal:
$$AA^T = U\Sigma ^2 U^T$$
Then multiplied by $U$ from the right:
$$AA^TU = U\Sigma ^2$$
Which is exactly the definition of $U$ as eigenvectors of $AA^T$ and $\Sigma ^2$ as its eigenvalues.
A: The product matrices $\mathbf{A}^{*} \mathbf{A}$ and $\mathbf{A} \mathbf{A}^{*}$ are symmetric. Symmetric matrices can be diagonalized by unitary matrices. The column vectors of the unitary matrices are the eigenvectors. The diagonal matrix is composed of the eigenvalues.
