Is every invertible matrix over the complex numbers diagonalizable?
Some relevant facts:
- A $n\times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors
- $\mathbb{C}$ is algebraically closed, and so every degree $n$ polynomial has $n$ (not necessarily distinct) roots (including the characteristic polynomial)
Sorry if this question is simple and I'm just not seeing it :)