Eigenvalues of $XX^T$

Question

I have a matrix $X$ that has $n\times m$ dimension. I want to compute the eigenvalues of $X^{T}X$ and $XX^{T}$. What is the relationship of the eigenvalues and eigenvector of $X^{T}X$ and $XX^{T}$? They have different dimensions so intuitive one matrix has more eigenvalues than the other, but I saw a result that says that there are the same non negative number of eigenvalues both matrices. How can I prove ir or realize it?

Answer 1

Start with the eigenvalue equation of $AB$:

$$ABv = \lambda v \implies B(AB)v = \lambda Bv \implies BA(Bv)=\lambda (Bv).\tag{*}\label{star}$$

If you substitute $w = Bv$ you will obtain the eigenvalue equation for $BA$. Hence, the eigenvalues of $AB$ are the same as the eigenvalues of $BA$. As @Lord Shark the Unknown said the larger will have additional $0$ eigenvalues. Let us assume the small product has $n$ eigenvalues. Then the larger one is forced to have the same $n$ eigenvalues. But because it is larger the only way to have additional eigenvalues is to have zero eigenvalues. If the eigenvalues were not zero, they would violate the equation \eqref{star}.