If one is looking at the characteristic polynomial of the $m \times m$ dimensional matrix $Z^TZ$ then apparently the coefficient of $(-1)^{m-k}$ in it can be written as, $\sum_{U \subset [m], V \subset [m], |U|=|V| = k}det^2(Z_{U,V})$ where by $Z_{U,V}$ one means the sumbatrix of $Z$ corresponding to the $k$ rows and columns of it given by the index sets $U$ and $V$ respectively.
Can someone kindly give a reference to a proof of this? (Or type in the proof if its short!)
I'll prove that the coefficient of $(-\lambda)^{m-k}$ in the characteristic polynomial of $AB$ is given by $\sum_{|U|=|V|=k} \det A_{U,V}\det B_{V,U}$. By the Cauchy-Binet formula, this sum is equal to $\sum_{|U|=k} \det((AB)_{U,U})$.
This latter quantity can be proved to be equal to the coefficient of $(-\lambda)^{m-k}$ in the characteristic polynomial of $AB$ by a combinatorial argument: to obtain a term of degree $m-k$ in the Laplace expansion of the characteristic polynomial, you need to select (in all possible ways) a permutation with at least $m-k$ fixed points, take $-\lambda$ terms from those fixed points and non-$\lambda$ terms from the other $k$. Summing over all possible subsets $[m]\setminus U$ of $m-k$ entries, you get that sum.
There must be a more elegant way to make this last argument, but that should be the idea at least.
