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I don't believe this matrix is diagonalizable, but I wanted to make sure.

Here's the matrix:

$A=\begin{bmatrix} 4 & 0 & 0 \\2 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$

A nice thing about this matrix is that it is diagonal, so the eigenvectors are $\lambda_1 = 4$, and $\lambda_2 = 2$ with algebraic multiplicity $2$.

The associated eigenvector for $\lambda_1$ is $x_1 = (1,1,1)$. When trying to solve for the eigenvector of $\lambda_2$, one gets the reduced matrix

$C = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Here, there is only the trivial solution, and that can't be an eigenvector due to definition. Based on that, can I rightly say that this matrix is not diagonalizable?

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I bet you meant "A nice is triangular..." , and not "diagonal", am I right? – DonAntonio Dec 2 '12 at 20:41
Ha, I didn't catch that! Yes, I meant triangular. – Kevin Dec 2 '12 at 20:45
@Tim: I am not sure what exactly do you mean by a "reduced matrix", but $A$ does have an eignvector for $\lambda_2$: $v=(0,0,1)^T$. – user1551 Dec 2 '12 at 21:06
By reduced matrix, I mean a row equivalent matrix under GJE. – Kevin Dec 2 '12 at 22:02
up vote 2 down vote accepted

Yes, your argument works. You can also remark that since the minimal polynomial of the matrix is the same as its characteristic one, i.e. $(x-4)(x-2)^2\,$ , then as the min. pol. is not a product of different linear factors the matrix isn't diagonalizable.

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I didn't think of that, thanks. – Kevin Dec 2 '12 at 20:44

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