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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 $A$?

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    See also https://math.stackexchange.com/a/336078/ for an outline of the proof.2017-07-18

3 Answers 3

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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.

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    @Leth This is a well know fact which you can prove by yourself by using the definition of determinant. See here https://math.stackexchange.com/questions/1148302/effect-of-row-operations-on-determinant-for-matrices-in-row-form?noredirect=1&lq=1 for pointers.2017-07-19
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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.

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