Diagonalize the matrix or explain why it cant be diagonalized
$A=\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}$
Hint: One eigenvalue is $λ=9$
So, i began the problem by finding the characteristic polynomial which was
$λ^3-7λ^2-15λ-27$
using long division i got $(λ-9)(λ^2+2λ+3)$
so i used quadratic formula and got
$λ=-1+i\sqrt{2}$ and $λ=-1-i\sqrt{2}$ and the given $λ=9$
I've never seen a problem with imaginary numbers when finding the eigenvectors so i'm wondering if that means it can't be diagonalized?