# 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 – Dilawar May 12 '12 at 3:39
• – Dilawar May 12 '12 at 3:46
• www.maa.org/sites/default/files/Louis_L49930._Pennisi.pdf – user26857 May 16 '16 at 7:46
• 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. – darij grinberg Jul 18 '17 at 12:50
• See also math.stackexchange.com/a/336078 for an outline of the proof. – darij grinberg Jul 18 '17 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. – onurcanbektas Jul 18 '17 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. – Dilawar Jul 19 '17 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. – Duchamp Gérard H. E. Feb 5 at 8:36

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: $$p^{(i)}(0)=i!\sum_{|J|=n-i}A[J]$$