For this problem, I was given the following matrix:
$$A=\begin{pmatrix} 3 & 0 & 0 \\ 2 & 4 & 0 \\ 1 & 1 & 8 \end{pmatrix}$$
One of the eigenvalues found, is $\lambda=8$. To find the eigenspace, I solved the following equations:
$$ (\lambda I-A)v=0 \implies \begin{pmatrix} 5 & 0 & 0 \\ -2 & -4 & 0 \\ -1 & -1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
This leads to $5a=0 \implies a=0 \implies -2*0-4b=0 \implies b=0$. These equations do not give further information about $c$. My question here is, how to construct the eigenspace from this? At first I thought it would look like the following:
$$ E_8(A)=\left\{\left( \begin{smallmatrix} 0 \\ 0 \\ c \end{smallmatrix}\right): c\in\mathbb{R} \right\}= span\left( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix}\right) $$ Since $a=b=0$ and no matter what is picked for $c$, the equations still hold. However, by intuition and looking at $N(T)$, being the null-space of the linear transformation corresponding to the matrix $(\lambda I-A)$, I'm guessing that the eigenspace is two-dimensional.
I'm not sure how to go on from here as the only ''two-dimensional'' space I can think of by this, is a space consisting of $(0,0)$ which is not even two-dimensional.
Your matrix has 3 distinct eigenvalues ($3,4$, and $8)$, so it can be diagonalized and each eigenspace has dimension $1$.
By the way, your system is wrong, even if your final result is correct. The right linear system is $\begin{pmatrix} 5 & 0 & 0 \\ 2 & -4 & 0 \\ 1 & 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \\ c\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$
You send get $a=0$, $b=0$ and $c$ arbitrary, which yields that your eigenspace is generated by $\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$.