In class, my professor mentioned the following technique as a fast way to compute the eigenpairs of a fat matrix (a matrix where the number of columns is much greater than the number of rows).
For an input matrix $X$ of dimensions $m\times n$, where $n$ is very large, compute $B=XX^T$.
Compute the SVD of $B$
Am I correct in understanding that by finding the eigenpairs of a matrix $B=XX^T$, we also find the eigenpairs of $X$; that is, the eigenpairs of $X$ and $XX^T$ are the same?
If so, why is this the case?