# Coefficients of characteristic polynomial of a matrix

For a given $n \times n$-matrix $A$, and $J\subseteq\{1,...,n\}$ let us denote by $A[J]$ its principal minor formed by the columns and rows with indices from $J$.

If the characteristic polynomial of $A$ is $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then why $$a_k=(-1)^{n-k}\sum_{|J|=n-k}A[J],$$ that is, why is each coefficient the sum of the appropriately sized principal minors of $A$?

• Found something useful .. www.mcs.csueastbay.edu/~malek/Class/Characteristic.pdf May 12, 2012 at 3:39
• May 12, 2012 at 3:46
• This follows from Corollary 5.161 in my Notes on the combinatorial fundamentals of algebra, version of 25 May 2017. Just mentioning this for the sake of completeness; I'm sure you don't want to read my proof (which is an unenlightening orgy of notation, with nothing interesting going on other than repeated applications of multilinearity), but it might be comforting to know it exists. Jul 18, 2017 at 12:50
• See also math.stackexchange.com/a/336078 for an outline of the proof. Jul 18, 2017 at 12:51

Use the fact that $\begin{vmatrix} a & b+e \\ c & d+f \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a & e \\ c & f \end{vmatrix}$

We can use this fact to separate out powers of $\lambda$. Following is an example for $2 \times 2$ matrix. $$\begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix} = \begin{vmatrix} a & b \\ c & d-\lambda \end{vmatrix} + \begin{vmatrix} -\lambda & b \\ 0 & d-\lambda \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + %% \begin{vmatrix} a & 0 \\ c & -\lambda \end{vmatrix} + %% \begin{vmatrix} -\lambda & b \\ 0 & d \end{vmatrix} + \begin{vmatrix} -\lambda & 0 \\ 0 & -\lambda \end{vmatrix}$$

This decompose $det$ expression into sum of various powers of $\lambda$.

Now try it with a $3 \times 3$ matrix and then generalize it.

• I couldn't understand how you went from LHS to RHS in the first equal sing in the$2\times 2$ matrix example. I mean I can see the equality by just calculating the determinants, but I couldn't get the method you used while separating the determinants.
– Our
Jul 18, 2017 at 15:17
• @Leth This is a well know fact which you can prove by yourself by using the definition of determinant. See here math.stackexchange.com/questions/1148302/… for pointers. Jul 19, 2017 at 7:39

One way to see it: $A:V\to V$ induces the (again linear) maps $\wedge^k A:\wedge^k V\to \wedge^k V$. Your formula (restated in an invariant way, i.e. independently of basis) says that $$\det(xI-A)=x^n-x^{n-1}\operatorname{Tr}(A)+ x^{n-2}\operatorname{Tr}(\wedge^2 A)-\cdots(*)$$ We can conjugate $A$ so that it becomes upper-triangular with diagonal elements $\lambda_i$ ($\lambda_i$'s are the roots of the char. polynomial). Now for upper triangular matrices the formula $(*)$ says that $$(x-\lambda_1)\cdots(x-\lambda_n)=x^n-x^{n-1}(\sum\lambda_i)+x^{n-2}(\sum\lambda_i\lambda_j)-\cdots$$ which is certainly true, hence $(*)$ is true.

• This is highbrow but doesn't explain the combinatorial equality of the OP. Feb 5, 2019 at 8:36
• As for why $\operatorname{tr}(\Lambda^k A)$ is the sum of principal $k\times k$ minors of $A$, see math.stackexchange.com/q/1604461. The references at mathoverflow.net/a/372497 are also useful, and math.stackexchange.com/q/23899 discusses an interesting extension of this result. Jan 16, 2021 at 2:29

$$\newcommand\sgn{\operatorname{sgn}}$$ I learned of the following proof from @J_P's answer to what effectively is the same question. It arises from expanding the usual definition $$\det A=\sum_{\sigma\in S_n}\sgn\sigma\prod_{1\le k\le n}A_{k,\sigma(k)}$$, and deserves to be more well-known than it currently is.

Let $$[n]:=\{1,\dots,n\}$$, and write $$\delta_{i,j}$$ for the Kronecker delta, which is equal to $$1$$ if $$i=j$$, and is $$0$$ otherwise. Note that $$\prod_{1\le k\le n}(a_k+b_k)=\sum_{C\subseteq[n]}\prod_{i\in C}a_i\prod_{j\in[n]-C}b_j$$, since every term in the expansion on the left hand side will choose from each expression $$(a_k+b_k)$$ either $$a_k$$ or $$b_k$$, and so we may sum over all possible ways $$C$$ of choosing the $$a_k$$ terms. We compute \begin{align*} \det(tI-A) &=\sum_{\sigma\in S_n}\sgn\sigma\prod_{1\le k\le n} (t\delta_{k,\sigma(k)}-A_{k,\sigma(k)})\\ &=\sum_{\sigma\in S_n}\sgn\sigma\sum_{C\subseteq[n]} \prod_{i\in C}(-A_{i,\sigma(i)})\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}\\ &=\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_n}\sgn\sigma \prod_{i\in C}A_{i,\sigma(i)}\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}. \end{align*} For fixed $$C\subseteq[n]$$ and $$\sigma\in S_n$$, the last product $$\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}$$ vanishes unless $$\sigma$$ fixes the elements of $$[n]-C$$, in which case the product is just $$t^{n-|C|}$$. So we need only consider the contributions of the permutations of $$C$$ in our sum, by thinking of a permutation $$\sigma\in S_n$$ that fixes $$[n]-C$$ as a permutation in $$S_C$$. The sign of this permutation considered as an element of $$S_C$$ remains the same, as can be seen if we consider the sign as $$(-1)^{T(\sigma)}$$, where $$T(\sigma)$$ is the number of transpositions of $$\sigma$$. We thus have \begin{align*} \sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_n}\sgn\sigma \prod_{i\in C}A_{i,\sigma(i)}\prod_{j\in[n]-C}t\delta_{j,\sigma(j)} &=\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_C}\sgn\sigma \prod_{i\in C}A_{i,\sigma(i)}t^{n-|C|}\\ &=\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\sum_{\sigma\in S_C}\sgn\sigma \prod_{i\in C}A_{i,\sigma(i)}. \end{align*} The term $$\sum_{\sigma\in S_C}\sgn\sigma\prod_{i\in C}A_{i,\sigma(i)}$$ is precisely the determinant of the principal submatrix $$A_{C\times C}$$, which is the $$|C|\times|C|$$ matrix with rows and columns indexed by $$C$$, and so \begin{align*} \sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\sum_{\sigma\in S_C}\sgn\sigma \prod_{i\in C}A_{i,\sigma(i)} &=\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\det(A_{C\times C})\\ &=\sum_{0\le k\le n}\sum_{\substack{C\subseteq[n]\\|C|=k}}(-1)^kt^{n-k} \det(A_{C\times C})\\ &=\sum_{0\le k\le n}t^{n-k}\left((-1)^k \sum_{\substack{C\subseteq[n]\\|C|=k}} \det(A_{C\times C})\right)\\ &=\sum_{0\le k\le n}t^k\left((-1)^{n-k} \sum_{\substack{C\subseteq[n]\\|C|=n-k}} \det(A_{C\times C})\right). \end{align*}

• I've been googling this question and this is the first answer that made perfect sense to me, thanks. Nov 24, 2021 at 20:07

Here's another way by using Taylor's theorem.

Consider $\det (xI+A)$ as a polynomial $p(x)$, from Taylor's theorem we have that: $$p(x)=\sum_{i=0}^n\frac{p^{(i)}(0)}{i!}x^i.$$ Computing $p^{(i)}(0)$ will leads quikly to the conclusion.

How to compute $p^{(i)}(x)$ at $x=0$ ? Well, here's a trick:

For instance we compute $p'(0)$, go back to the determinant and replace the $x$ in the $k$th row by $x_k$, and using the total derivative. Then you'll find: $$p'(0)=\sum_{|J|=n-1}A[J].$$

And using induction we can show in general that: $$p^{(i)}(0)=i!\sum_{|J|=n-i}A[J]$$