# Characteristic Eq for Matrix problem

## Homework Statement

Where A is an n x n matrix and I is the n x n identity
In the expansion of det(x*I-A), show that the coefficient of x is equal to the sum from i = 1 to n of the determinant of the Aii minor. (where Aii = the submatrix of A formed by deleting row i and column i)

## Homework Equations

if X1, X2, ... Xn are the eigenvalues of A then the coefficient of x is equal to the sum from i = 1 to n of (product of eigenvalues)/Xi.

## The Attempt at a Solution

I have tried quite a few different ways of doing this but getting nowhere..
using the definition of a determinant as a sum of products over permuations of Sym (n) and then using the Leibnix rule for integrals I seem to find that (where a11 is the top left entry of A and so on)
d(det(x*I-A)/dx = sum (from i =1 to n) of det(x*I-A)/(x-aii)

The constant coefficient of this will be equal to the x coefficient of det(x*I-A) but the RHS only cancels directly to what I want if aii is an eigenvalue, which obviously isn't always the case. I tried to use row permutations but this obviously changes the Aii minors and so doesn't seem to work. Any ideas? I was trying to follow the proof on

http://www.sciencedirect.com/scienc...26cc582e2a0f1358d2f4b36931f4954e&searchtype=a

(sorry for the long link!) but the notation is a bit beyond me. Am I on the right track at all or is there a better way to attempt this? Sorry for not including full working but without the maths language its near impossible to be clear.
Thanks

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hunt_mat
Homework Helper
Can you prove it for when A is diagonal

Yes I understand for when it is diagonal but altho row/column operations of form (row i -> row i + d*row j) won't change the determinant of the matrix, they will change the determinant of the minors.. I don't see how to relate the diagonal form back to the original matrix.

hunt_mat
Homework Helper
How about for a diagonaliable matrix? That is the key here.

Hmmmn I'm still pretty clueless but I'll give it another go tomorrow! Thanks