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I have a couple of questions regarding diagonalizable matrices. I think I know the answers, but I wish to confirm.

1) Is every triangular matrix also diagonalizable? I think not. The eigenvalues are the elements of the diagonal, and if one of them repeat itself, the matrix may not be diagonalizable.

2) Any matrix with distinct eigenvalues is diagonalizable. I think so. If the eigenvalues are distinct, then the algebraic multiplicity is $1$, meaning that it has to equal the geometric multiplicity.

Am I correct in both? I would appreciate it if you correct me if I am wrong with either explanation or conclusions. Thank you.

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    $\begingroup$ You are correct about both 1 and 2, but I think that a concrete counterexample would be a better explanation in 1. $\endgroup$ – heptagon Feb 28 '17 at 13:44
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1) You are right, this is not true.

Take for instance $$M=\begin{pmatrix}1 & -1 \\ 0 & 1 \end{pmatrix}$$ which is not diagonalizable (because the only eignen value is $1$ and it would be an homothethy and it is not the case).

But is is trivially triangular.

2) You are also correct, because the minimal polynomial factors completely with only simple roots.

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