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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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    Found something useful .. www.mcs.csueastbay.edu/~malek/Class/Characteristic.pdf2012-05-12
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    Also http://books.google.co.in/books?id=ULMmheb26ZcC&pg=PA195&dq=coefficients+of+characteristic+polynomial&hl=en&sa=X&ei=q9utT7PpFI3JrAeWpcWUBA&redir_esc=y#v=onepage&q=coefficients%20of%20characteristic%20polynomial&f=false2012-05-12
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    www.maa.org/sites/default/files/Louis_L49930._Pennisi.pdf2016-05-16
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    This follows from Corollary 5.161 in my [*Notes on the combinatorial fundamentals of algebra*, version of 25 May 2017](https://github.com/darijgr/detnotes/releases/tag/2017-05-25). 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.2017-07-18
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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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    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.2017-07-18
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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: $$p^{(i)}(0)=i!\sum_{|J|=n-i}A[J]$$