# Coefficients of the characteristic equation

1. Apr 3, 2012

### spaghetti3451

let's say you are given the following matrix T:

T11 T12 . . . T1n
T21 T22 . . . T2n
. . .
. . .
. . .
Tn1 Tn2 . . . Tnn

You want to find the characteristic equation for the matrix, which is
$C_{n} \lambda^{n} + C_{n-1} \lambda^{n-1} + ... + C_{1} \lambda + C_{0} = 0.$

Now, $C_{n} = (-1)^{n}$, $C_{n-1} = (-1)^{n-1} Tr(T)$, and $C_{0} = det(T)$.

I have been having a pretty hard proving these three formulae. The first one looks intuitively obvious and I think I have a fairly good understanding of why the second one has the structure that it has. But the complexity of the matrix notation and the fact that there are so many elements in the matrix is halting my attempts to precisely determine each of the formulae from the matrix. I would appreciate any help. Thanks.

Last edited: Apr 3, 2012
2. Apr 3, 2012

### DonAntonio

The characteristic polynomial, not equation (unless equalled to zero or some other number) is the determinant of $T-xI$ .

If we also know that the determinant is the sum of products of n numbers, each one of which is formed with exactly one element from each row and each column of the matrix, we get at once that one of such products in the above determinant is $(-x)(-x)\cdot...\cdot (-x)=(-1)^nx^n$ , and there is no other product as above with n x's, so this last is the greatest degree term in the pol., and you get $C_n$ .

Now you try to deduce the other two.

DonAntonio

3. Apr 7, 2012

### ajkoer

Co = Det (T) is easy. The last term Co is always the product of the roots. The roots of the characteristic equation for the matrix are the eigenvalues. As the product of the eigenvalues is equal to the Determinant, we are done. Proof of the last statement:

Det (T) = Det (P'EP) = Det(P'PE) = Det (I*E) = Det (E) = e1*e2*...

where E is a diagonal matrix with the eigenvalues on the diagonal of matrix T and P'P = I.

Similarly, the second term is always the minus sum of the roots. Note, rescaling the term becomes C(n-1)/C(n) = -1 as was required.

Last edited: Apr 7, 2012
4. Apr 7, 2012

### morphism

One minor difficulty with this approach is that the OP didn't specify the field over which they're working, so one might have to enlarge it so that it contains the eigenvalues of T, and then one has to prove that det and Tr once taken over this enlarged field will give the same values as over the original field.

A more serious problem, however, is that your proof seems to assume that T is diagonalizable. This issue really needs to be addressed. (Suggestion: If we're working over a field that contains the eigenvalues of T then, although T may not be diagonalizable over this field, it's still possible to find a matrix P such that $PTP^{-1}=E$ is upper triangular. This would salvage your argument.)