Is it possible for an eigenspace to have dimension $0$? 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.
 A: 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)$\le n$.
A: Following the definition, $\lambda$ is an eigenvalue of the matrix $A$ if there exists a non-zero vector $v$ such that:
$$Av = \lambda v.$$
The definition itself assures that, if $\lambda$ is an eigenvalue, then there must be also an eigenvector $v$. The presence of at least one eigenvector implies that the eigenspace relative of $\lambda$ has at least dimension equal to $1$.
You cannot define an eigenvalue without an eigenvector, and viceversa.
A: It is a matter of convention. What everybody should agree on is that $\lambda$ being an eigenvalue of$~A$ means that $\dim(\ker(A-\lambda I))>0$, so the dimension of the eigenspace associated to an eigenvalue is never$~0$. However if the dimension in that formula is$~0$, so if $\lambda$ is not an eigenvalue, then one could still agree that $\ker(A-\lambda I)$ may be called the eigenspace $E_\lambda$ of$~A$ for (the non-eigenvalue)$~\lambda$. This may seem a bit weird, but there are many occasions where it is a convenience to be able to talk about $E_\lambda$ for any scalar$~\lambda$, without first having to ensure that $\lambda$ is an eigenvalue. For instance, if $A$ is a projector (so $A^2=A$) then it is always true that the whole space decomposes as $E_0\oplus E_1$, though it might happen that one of $E_0$ and $E_1$ has dimension$~0$ (namely if $A=I$ respectively $A=0$); this statement would be more complicated to make if it were forbidden to mention $E_\lambda$ unless $\lambda$ was actually an eigenvalue.
A: It doesn't imply that dimension 0 is possible. You know by definition that the dimension of an eigenspace is at least 1. So if the dimension is also at most 1 it means the dimension is exactly 1. It's a classic way to show that something is equal to exactly some number. First you show that it is at least that number then that it is at most that number.
