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)
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
(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.