Coefficient of det($xI+A$)

Question

I want to know how to prove the fact that the coefficient of $x^k$ in $\det(xI+A)$ is the sum of all $(n-k)\times (n-k)$ submatrices of $A$, an $n \times n$ matrix.

Answer 1

This is true, except you only want to consider the determinants of the principal submatrices: i.e. the $[A_{i,j}]_{i,j \in I}$ for some set $I$ (the same one for $i$'s and $j$'s).

However you approach the issue, I think it leads to some mundane/vague transformations. You can, for instance, write out the formula for $\det(xI + A)$ as a sum over permutations, and open the brackets of the terms like $(x+A_{i,i})$. Now, to get $x^k$, you need to choose precisely $k$ occurrences of $x$: this means, you have to fix $k$ indices $i_1,i_2,\dots,i_k$, and the corresponding permutation $\pi$ has to map $i_l$ to $i_l$ (for $l=1,2,\dots,k$). On the remaining indices, you need not to get any $x$'s, which means you can essentially forget the $x$'s were there in the first place, and just consider the entries of $A$. If you sum over all possible ways in which you get $x^k$ out of the indices $i_1,i_2,\dots,i_k$, you recover the formula for the determinant of the matrix $[A_{i,j}]_{i,j \in I}$ where $I = \{1,2,\dots,n\} \setminus \{i_1,i_2,\dots,i_k\}$. Sum over all choices of $i_1,i_2,\dots,i_k$ to get your claim.

By the way, I do realise it is terribly vague, but I believe it can be made to work. I would welcome a nicer reasoning.