Every $n\times n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).
In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.
In the latter case, I can see two possibilities:
There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.
There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.
Is one of these two possibilities always true, for all complex $n\times n$ matrices, or could either possibility be true depending on the matrix in question?
Is there a proof of this, or perhaps a counter example?