We know that eigenvectors to different eigenvalues are linearly independent. So the sum of dimensions cannot be larger than the dimension of $F^n$, which is $n$. (Simply because any subspace of $F^n$ has dimension at most $n$.)
The inequality can be strict, for example, $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has only one eigenvalue, it has eigenspace of dimension one. (You can take any matrix which is not diagonalizable.)