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

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?

• This does or does not mean it can't be diagonalized, depending on whether you're supposed to be allowed to use complex eigenvectors or not. We can't say, you need to ask the person in charge... – David C. Ullrich Aug 7 '15 at 16:51
• Diagonalization is a field dependent concept. It is diagonalizable over $\mathbb C$, but not over $\mathbb R$. You can use the same method to diagonalize over $\mathbb C$ that you'd use to diagonalize over $\mathbb R$. Scalars will be scalars... – Git Gud Aug 7 '15 at 16:52
• okay thank you, i just wanted to check before i continued. – Charlene Aug 7 '15 at 16:53
• @Surb yes, thank you. – Charlene Aug 7 '15 at 16:57
• There's something i don't get. Why was that hint given by whoever made this problem? Since you ended up finding that eigenvalue easily after finding the characteristic polynomial. Was there a simpler way to find the other 2 by assuming that hint correct from the go? – Dleep Aug 7 '15 at 17:03

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
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.