Originally Posted by

**bkbowser** Yes that I think is true. I am having trouble showing it though.

So I know the eigenvalues go on the diagonal. And I guess it's no problem to show that each of $1$, $\omega$, and $\omega^2$ solve $1-\lambda^3$. So then I think I'm done and have only to write $\dots \begin{pmatrix}1&0&0\\0&w&0\\0&0&w^2\end{pmatrix} \blacksquare$

Most proofs involving "Jordanizing" some transformation involve some business about the minimal polynomial but that seems irrelevant as there are 3 eigenvalues hence the matrix is diagonable. Or is there more?