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Consider just the characteristic polynomial of the following matriz and justify that is is non diagonalizable:

$$ \begin{bmatrix} 0 & 0 & 4 \\ 3 & 2 & -1 \\ 1 & 0 & -4 \\ \end{bmatrix} $$

So I calculated the characteristic polynomial $$(2-x)(x^2 + 4x - 4)$$ and verified the matrix has three eigenvalues: $$2, (2\sqrt2 - 2), (-2\sqrt2 - 2)$$

Is the matrix non diagonalizable because the eigenvalues are non integers? Or is there another reason?

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    Perhaps the last column is $(-4,-1,4)$. – 2017-01-19
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    Just checked with the book, the matrix I posted is correct! – 2017-01-19
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    Perhaps a typo. With $(4,-1,-4)$, the matrix is diagonalizable as it has $3$ real and distinct eigenvalues. Try the exercise with this new column... – 2017-01-19

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The eigenvalues of $A=\begin{bmatrix} 0 & 0 & 4 \\ 3 & 2 & -1 \\ 1 & 0 & -4 \\ \end{bmatrix}$ are $2$ and $-2\pm\sqrt{2}$. So, $A$ has three simple eigenvalues in $\mathbb{R}$ and is diagonalizable over $\mathbb{R}$ but not over $\mathbb{Q}$ (has irrational eigenvalues).

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    Just corrected the eigenvalues. Still, it's strange because my book doesn't specify if it's over $$\mathbb{R} or \mathbb{Q}$$ – 2017-01-19
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    Well, the fact is that $A$ is diagonalizable over $\mathbb{R}$ and not over $\mathbb{Q}.$ We can't change that absolute truth. :) – 2017-01-19
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    Most probably they meant matrix over the reals $\;\Bbb R\;$ . I don't know one single book in linear algebra that singles out rational matrices as something special, but real matrices are very common. Thus, the matrix is diagonalizable as it has three different eigenvalues and it is a $\;3\times 3\;$ matrix. – 2017-01-19
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    Probably as a "non written" agreement, but writing $A\in\mathbb{K}^{n\times n}$ and specifying $\mathbb{K}$ we would avoid dichotomies. – 2017-01-19