Eigenvalues of a matrix multiplied by its transpose

Question

I recall being told that the eigenvalues of the matrix formed by multiplying a matrix by its transpose are the squares of the eigenvalues of the original matrix. Is that true for any matrix? Surely not.

Answer 1

No. Take $A=\left(\begin{smallmatrix}1&2\\3&2\end{smallmatrix}\right)$, for instance. Then the eigenvalues of $A$ are $4$ and $-1$. But the eigenvalues of $A^t.A$ don't even belong to $\mathbb Z$; they are $9\pm\sqrt{65}$.

Answer 2

No they're definitely not. Here's an easy counterexample:

$$ A=\left( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} \right) $$ This matrix has eigenvalues $1$ and $0$ but $$ A^T A = \left( \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} \right) $$ which has eigenvalues $2$ and $0$.

It is true if $A$ is a normal matrix (i.e. $A^T A = AA^T$) though.