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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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    By reduced matri$x$, I mean a row equivalent matri$x$ under GJE.2012-12-02

1 Answers 1

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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.2012-12-02