# Is every invertible matrix over an algebraically closed field diagonalisable?

In $\Bbb{R}$ the only invertible matrices (I can think of) that are not diagonalisable are those which stand for a rotation, but in $\Bbb{C}$ this shouldn't be a problem anymore, since rotations can be expressed via a multiplication with a complex scalar just fine.

So are there any invertible matrices over $\Bbb{C}$ that are not diagonalisable?

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$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ –  Daniel Fischer Apr 23 '14 at 19:13
I see my error in reasoning now: I still had orthogonal matrices in mind and the fact that they consist of rotations and mirroring(?)... thanks for clarifying! –  Peter Apr 23 '14 at 19:18
Spiegelung = reflection –  Daniel Fischer Apr 23 '14 at 19:22
Thanks a lot! Do you want to post your comment as an answer, so I can mark this question as answered? –  Peter Apr 23 '14 at 19:26
At the moment, I lack the energy to whip up a full answer. Anybody is invited to have a go. –  Daniel Fischer Apr 23 '14 at 19:27

As Daniel Fischer kindly pointed out in the comments there is a counterexample that works in $\Bbb{C}$ just like it does in $\Bbb{R}$: