Diagonalize the matrix A or explain why it can't be diagonalized

Question

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?

Answer 1

A matrix $A \in M_{n\times n}(\mathbb F)$ is diagonalizable iff:

A. The characteristic polynomial has all its roots in $\mathbb F$ and

B. The algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity.

Having said that, we have that every eigenvalue is simple (that means B is satisfied, in any case).

If we consider our matrix $A \in M_{3\times 3}(\mathbb C)$ then it is diagonalizable. However, if we consider our matrix $A \in M_{3\times 3}(\mathbb R)$, then it is not diagonalizable.