Let $$A = \begin{pmatrix} -1 & 2 & -2 \\ 5 & -1 & 6 \\ 6 & -2 & 7 \end{pmatrix}$$.
The matrix $A$ has eigenvalues $5, \pm i$. Now I know that $A$ is only diagonalisable over $\mathbb{C}$, and not diagonalisable over $\mathbb{R}$, since we only have one real eigenvalue. But it says that $A$ is not diagonalisable, just because it has one eigenvalue. Isn't this the sole reason?
Since the eigenspaces over $\mathbb{R}$ only have one dimension, then it's not diagonalisable.
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Twenty-six colours
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1You've got it right. They are only considering diagonalizability over $\mathbb R$. – Jack's wasted life May 14 '17 at 06:36
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So I have the correct reason? I think the notes said that that wasn't the only reason – Twenty-six colours May 14 '17 at 06:39
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1That's about it. Eigenspaces over $\mathbb R$ must have three dimensions in total for it to be diagonalizable. – Jack's wasted life May 14 '17 at 06:41