Is the sum of the dimensions of the eigenspaces of an $n\times n$ matrix necessarily equal to $n$?

but I was advised to simply show that $1\leq\mathrm{dim}(\text{sum of all the eigenspaces})\leq n$ by verifying it for a $2\times 2$ matrix

how can we do that?

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.)

@MartinSleziak Thanks!