How can I find the dimension of the eigenspace?

By definition, an eigenvector $v$ with eigenvalue $\lambda$ satisfies $Av = \lambda v$, so we have $Av-\lambda v =Av - \lambda I v = 0$, where $I$ is the identity matrix. Thus, $$(A-\lambda I)v = 0,$$ and $v$ is in the nullspace of $A-\lambda I$.

Since the eigenvalue in your example is $\lambda = 8$, to find the eigenspace related to this eigenvalue we need to find the nullspace of $A - 8I$, which is the matrix $$\left[\begin{array}{cc}1 & -1 \\ 1 & -1 \\ \end{array} \right].$$ We can row-reduce it to obtain $$\left[\begin{array}{cc} 1 & -1 \\ 0 & 0 \\ \end{array} \right].$$ This corresponds to the equation $$x-y = 0,$$ so $x = y$ for every eigenvector associated to the eigenvalue $\lambda = 8$. Therefore, if $(x,y)$ is an eigenvector, then $(x,y) = (x,x) = x(1,1)$, meaning that the eigenspace is $$W=[(1,1)],$$ and its dimension is $1$.

The dimension of the eigenspace is given by the dimension of the nullspace of $A - 8I = \left(\begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right)$, which one can row reduce to $\left(\begin{matrix} 1 & -1 \\ 0 & 0 \end{matrix} \right)$, so the dimension is $1$.

Note that the number of pivots in this matrix counts the rank of $A-8I$. Thinking of $A-8I$ as a linear operator from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$, the dimension of the nullspace of $A$ is given by $\dim(\mathbb{R}^{2}) - \mathrm{rank}(A) = 2 - 1 = 1$ by the so-called rank-nullity theorem. Of course, one can be more explicit: it is straightforward to see that the nullspace of $A - 8I$ is spanned by the vector $(1, 1)$, and hence has dimension $1$.