1. ## Eigenvalue/Eigenvector

I am loosing my mind. I cannot seem to factor the ^3-4^2-10-12.

My problem is to verify that $e=[1,1,1]^T$ is an eigenvector of A=(1,2,3), (3,2,1), (2,3,1).

det(-AI) = (-1)(-2)(-1)+(-2)(-1)(-2)+(-3)(-3)(-3)-(-2)(-2)(-3)-(-1)(-3)(-1)-(-1)(-2)(-3)
= ^3-4^2-10-12
= ???

2. ## Re: Eigenvalue/Eigenvector

since you are given the eigenvector, just calculate:

$\begin{bmatrix}1&2&3\\3&2&1\\2&3&1\end{bmatrix} \begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}6\\6\\6\end{bmatrix}$

this tells you that 6 is an eigenvalue.

now divide the characteristic polynomial by λ-6.