Dimension of the corresponding eigenspace?

The dimension is two. Note that the vectors $ u=\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array} \right] $ and $v= \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ \end{array} \right] $ are in the null space of $A-I_4=\begin{bmatrix} 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}$, i.e. $$Au=u\mbox{ and } Av=v.$$ So $u$ and $v$ are eigenvectors corresponding to the eigenvalue $1$. In fact, the form a basis for the null space of $A-I_4$. Therefore, the eigenspace for $1$ is spanned by $u$ and $v$, and its dimension is two.

In general, the eigenspace of an eigenvalue $\lambda$ is the set of all vectors $v$ such that $Av=\lambda v$. This also means $Av-\lambda v=0$, or $(A-\lambda I)v=0$. Hence, you can just calculate the kernel of $A-\lambda I$ to find the eigenspace of $\lambda$.