Finding the nullspace of a square matrix I wish to find the nullspace of $(A-2I)$, where $A$ is given below, as part of finding the kernel of a linear transformation.

And the answer provided is 

I am able to obtain the vector on the left, but I am not too sure how the vector on the right was acquired. I do not require an answer containing the exact computation. A brief explanation of the main steps involved is sufficient.
 A: The nullspace of $A-2I$ for the matrix provided (a.k.a. the eigenspace of $A$ since the associated eigenvalue is $2$) can be easily found by writing $A-2I$ in reduced row-echelon form and then solving $(A-2I)x=0$ for $x$. The basis set of $x$ is the nullspace requested.
Note: Since $\mathrm{rank}(A-2I)=2$ but $m=3$, there will be just one vector in the nullspace, i.e., $\dim(\mathrm{null}(A-2I))=m-\mathrm{rank}(A-2I)$.
A: You're computing the eigenspaces, the characteristic polynomial of the matrix is
$$
(3-X)(2-X)^2
$$
so the eigenvalues are $3$ and $2$. The eigenspace relative to $2$ is the null space of
$$
\begin{pmatrix}
3-2 & 0 & -1 \\
0 & 1-2 & 1 \\
0 & -1 & 3-2
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & -1 \\
0 & -1 & 1 \\
0 & -1 & 1
\end{pmatrix}
$$
and a Gaussian elimination gives
$$
\begin{pmatrix}
1 & 0 & -1 \\
0 & -1 & 1 \\
0 & -1 & 1
\end{pmatrix}
\to
\begin{pmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{pmatrix}
$$
so the equations for the eigenvectors are
$$
\begin{cases}
x_1=x_3\\[4px]
x_2=x_3
\end{cases}
$$
so the eigenspace has dimension $1$ and it is generated by
$$
\begin{pmatrix}1\\1\\1\end{pmatrix}
$$
The eigenspace relative to $3$ is the null space of
$$
\begin{pmatrix}
3-3 & 0 & -1 \\
0 & 1-3 & 1 \\
0 & -1 & 3-3
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 & -1 \\
0 & -2 & 1 \\
0 & -1 & 0
\end{pmatrix}
$$
A Gaussian elimination gives the equation $x_2=x_3=0$, so the eigenspace is generated by
$$
\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}
$$
