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

user26857
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user67458
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1 Answers1

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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 your approach the issue, I think, 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 $x$. That is, 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.

Our
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Jakub Konieczny
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  • It would be indeed nice if you explained it a bit in detail because I almost didn't understand anything at all. – Our Jul 18 '17 at 15:52
  • I don't think this works. Let $F(\sigma)$ be the set of fixed points of $\sigma$. Then\begin{align}\det(xI_n+A)&=\sum_{\sigma\in S_n}(-1)^\sigma\prod_{1\le k\le n}(x\delta_{k,\sigma(k)}+A_{k,\sigma(k)})\&=\sum_{C\subseteq[n]}\sum_{F(\sigma)=C}(-1)^\sigma\prod_{1\le k\le n}(x\delta_{k,\sigma(k)}+A_{k,\sigma(k)})\&=\sum_{C\subseteq[n]}\sum_{\sigma\in S_n,F(\sigma)=C}(-1)^\sigma\prod_{i\in C}(x+A_{i,i})\prod_{j\in[n]-C}A_{j,\sigma(j)}.\end{align}If $n=3$ only $F(23)={1}$ and we get $-(x+A_{1,1})A_{2,3}A_{3,2}$ which does not give us the minor we want. The remaining terms come from identity. – ho boon suan Jan 18 '21 at 05:06