# Having difficulty working out a Char. Pol. for Eigenvalues

1. May 31, 2015

### leo255

Given the following matrix:

2 3 0
3 2 4
0 4 2

I'm having a difficult time working out the characteristic polynomial. I used the shortcut that I saw on YT, where it is (I am using x instead of lambda) X^3 - (trace) X^2 + (A11+A22+A33) X - DET(A)

I got the following:

trace is just 2+2+2 = 6
A11+A22+A33 = (4-16) + (4-0) + (4-9) = -13
DET(A) = -42

This is giving me a characteristic polynomial of X^3 - 6X^2 - 13X + 42. Using synthetic division with the correct e-vals, it just doesn't come out right.

The correct roots/e-vals are -2, 2 and 4.

Thanks.

2. May 31, 2015

### RUber

You could go through the process of working it out without the shortcut,
$det \left[\begin{pmatrix}2 -x&3 &0\\3&2-x&4\\0&4&2-x\end{pmatrix}\right] = 0$
$(2-x)^3 -25(2-x) =0$
I don't get -2, 2, and 4 though. Are you sure you copied the matrix down right?