Given a linear map $L:V \to V$; $V$, $W$ vector spaces, the eigenspace $E$ associated with an eigenvalue $\lambda$ is that subspace of $V$ where $L$ acts as a scalar, i.e., $L$ acts by stretching vectors. The geometric multiplicity of $\lambda$ is the dimension of the subspace of $V$ (in the domain) where $T$ acts as a scalar, with scaling coefficient $\lambda$, i.e., $T$ when restricted to the eigenspace is a map given by $T(v)=\lambda v$.
Maybe the clearest example is that given by the identity map $I:V \to V$, $I(v)=v$, where $V$ is $n$-dimensional ($n < \infty$), with associated matrix the identity matrix. Here, the only eigenvalue is $1$, and you can easily see, e.g, by looking at the associated matrix ($M-\lambda I:=(I-1I)=0$ that the geometric multiplicity is $n$; this means that the identity acts on the whole of $V$ by scaling by $1$. Substituting the identity matrix for a scalar matrix (i.e. $a_{ii}=c$; $a_{ij}=0$; $i\neq j$) illustrates the same point, e.g., if our matrix is the matrix ($a_{ij}:a_{ii}=2$; $a_{ij}=0$ if $i \neq j$ ) then $M$ acts like a scalar for every vector.
At the other extreme, if our matrix represents a linear transformation of a rotation by an angle $\theta \neq 0, \neq 2n\pi$, then the eigenspace is $0$-dimensional, since each point will be sent to another point with the same radius (because it will send a point $p$ in a circle $C$ to another point $p'$ in the same circle), and no point will be rotated into itself. The eigenspace would then be $0$-dimensional, since only the dimension-zero subspace will be sent to itself.