I don't think it is possible because that entails that only the $\mathbf 0$-vector is in the eigenspace, but $\mathbf 0$ is not an eigenvector by definition.
However, my textbook says:
For an $n\times n$ matrix, if there are $n$ distinct eigenvalues, then all eigenspaces have dimension at most $1$.
which seems to imply that eigenspaces of dimension $0$ are possible.
Best Answer
By definition to any eigenvalues correspond at least one eigenvector thus for a n-by-n matrix for each eigenvalue $\lambda_i$ we have $1\le$ dim(eigenspace)$\le n$.