Eigenvalues of $A$ and $A A^T$ Let $A$ be a real $n\times n$ matrix. How are the eigenvalues of $A$ and $AA^T$ related?
What I have come up with so far is that if we let $\lambda_1,\ldots,\lambda_n$ denote the eigenvalues of $A$, then since
$$ \det(AA^T) = \det(A)^2 $$
we can deduce the product of the eigenvalues of $AA^T$ is the product of the eigenvalues of $A$ squared. 
This lead me to believe that $\lambda_1^2,\ldots,\lambda_n^2$ are the eigenvalues of $AA^T$, but I have not been able to prove it.
Any help is appreciated!
 A: The eigenvalues of $A A^T$ are nonnegative real numbers. Their square roots are called the singular values of $A$ (and $A^T$). Like the eigenvalues of $A$, they multiply out to the determinant. Nonetheless your statement usually fails. A rotation by $\pi/2$ in $\mathbb{R}^2$ shows the point easily enough: the eigenvalues are $\pm i$, their square is $-1$, but $A A^T=I$, whose eigenvalues are $1$. Requiring the eigenvalues to be real doesn't fix the matter, either.
You can get a relationship when $A$ is normal: in this case $A$ and $A^*$ (the conjugate transpose) commute, so they share eigenvectors. The eigenvalues of $A^*$ are the conjugates of the eigenvalues of $A$, however, even when $A$ was real to begin with. Thus for an eigenvector $x$ of $A^*$ with eigenvalue $\lambda$ you have $A A^* x = \lambda A x = \lambda \overline{\lambda} x = |\lambda|^2 x$. But even now taking the square root removes whatever complex phase that $\lambda$ may have had.
You could have proved the same thing from the assumption of unitary diagonalizability...but in fact the (generalized) spectral theorem tells us that this is equivalent to normality.
A: Recall that $A$ and $A^T$ have the same set of eigenvalues.
Since, for $\lambda \in \mathbb{R}$ we have that $Ax = \lambda x$ and $A^Tx = \lambda x$ we obtain
$$A^TAx = A^T(Ax) = A^T\lambda x = \lambda(A^Tx) = \lambda^2x$$
and similarly
$$AA^Tx = A(A^Tx) = A\lambda x = \lambda(Ax) = \lambda^2x$$
