Is every invertible matrix over an algebraically closed field diagonalisable?

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

\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}

My error in reasoning was that I still had orthogonal matrices and their trait of consisting of rotations and reflections in mind.