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

If the characteristic polynomial of A is $x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$, 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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Did you find a solution. I am also looking for an answer to this problem. – Dilawar May 12 '12 at 3:26
    
Found something useful .. www.mcs.csueastbay.edu/~malek/Class/Characteristic.pdf – Dilawar May 12 '12 at 3:39
    
Also books.google.co.in/… – Dilawar May 12 '12 at 3:46
    
www.maa.org/sites/default/files/Louis_L49930._Pennisi.pdf – user26857 May 16 at 7:46

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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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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