Find all eigenvalues and eigenvectors of this $3\times3$ matrix I'm reviewing for a Differential Equations exam and one of the questions in the practice exam asks me to find all the eigenvectors of $3 \times 3$ matrix $\mathbf{A}$ given that $1$ is an eigenvalue of the matrix.
The matrix $\mathbf{A}$ is
$$
\left(
\begin{array}{crr}
1 & -1 & 1 \\
1 & 0 & -1 \\
0 & -1 & 2 \end{array}
\right)$$
The solution mentions something about the eigenvectors satisfying
$$
 \left( \mathbf{A} - \lambda_{k} \mathbf{I} \right) v_{k} = \mathbf{0}
$$
but I'm not sure where this came from or why it's true. Could someone please explain?
 A: Problem statement
Resolve the eigensystem for 
$$
 \mathbf{A} =
%
\left[
\begin{array}{crr}
 1 & -1 & 1 \\
 1 & 0 & -1 \\
 0 & -1 & 2 \\
\end{array}
\right]
$$
Eigenvalues
The problem is succinctly formulated in the comment of @Bye_World. Start with
$$
  \mathbf{A} - \lambda \mathbf{I}_{3} =
\left[
\begin{array}{ccc}
 \boxed{1-\lambda}  & \boxed{-1} & \boxed{1} \\
 1 & -\lambda  & -1 \\
 0 & -1 & 2-\lambda  \\
\end{array}
\right]
$$
Define characteristic equation
$$
 p(\lambda) = \det \left( \mathbf{A} - \lambda \mathbf{I}_{3} \right)
$$
Construct determinant from minors
$$
%
\begin{align}
%
 \det \left( \mathbf{A} - \lambda \mathbf{I}_{3} \right)
%
&= \boxed{\left( 1 - \lambda \right)}
\left|
\begin{array}{cc}
 -\lambda  & -1 \\
 -1 & 2-\lambda  \\
\end{array}
\right|
%
- \boxed{-1}
\left|
\begin{array}{cc}
  1 & -1 \\
  0 & 2-\lambda  \\
\end{array}
\right|
%
+ \boxed{1}
\left|
\begin{array}{cc}
 1 & -\lambda  \\
 0 & -1 \\
\end{array}
\right| \\[3pt]
%
&= -\lambda  \left(\lambda ^2-3 \lambda +2\right) \\
%
&= -\lambda  \left(\lambda - 2 \right)\left(\lambda - 1 \right) \\
%
\end{align}
%
$$
The roots $p\left( \lambda \right) = 0$ are the eigenvalues.
The eigenvalue spectrum is
$$
 \lambda \left( \mathbf{A} \right) = \left\{ 2, 1, 0\right\}
$$
Eigenvectors
Solve $$ \left( \mathbf{A} -\lambda_{k} \mathbf{I}_{3} \right) u = 0$$
$ \lambda_{1} = 2$:
$$
%
\begin{align}
%
\left[
\begin{array}{c}
 -u_{1} - u_{2} + u_{3} \\
  u_{1} - 2 u_{2} - u_{3} \\
 -u_{2} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
 0 \\
 0 \\
 0 \\
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
v_{1} =
\left[
\begin{array}{c}
 1 \\
 0 \\
 1 \\
\end{array}
\right]
%
\end{align}
%
$$
$ \lambda_{2} = 1$:
$$
%
\left[
\begin{array}{c}
  - u_{2} + u_{3} \\
  u_{1} - u_{2} - u_{3} \\
 -u_{2} + u_{3} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
 0 \\
 0 \\
 0 \\
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
v_{2} =
\left[
\begin{array}{c}
 2 \\
 1 \\
 1 \\
\end{array}
\right]
%
$$
$ \lambda_{3} = 0$:
$$
%
\left[
\begin{array}{c}
  u_{1} - u_{2} + u_{3} \\
  u_{1} - u_{3} \\
 -u_{2} + 2u_{3} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
 0 \\
 0 \\
 0 \\
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
v_{3} =
\left[
\begin{array}{c}
 1 \\
 2 \\
 1 \\
\end{array}
\right]
%
$$
