Show that $\lambda$=1 as eigenvalue, find one corresponding eigenvector Here's the question:
$\lambda$

The typical formula I've seen is $(A-\lambda I)v = 0$ where A is the starting matrix, $\lambda$ is the eigenvalue, I is:
$$\begin{bmatrix}
1&0&0 \\
0&1&0 \\
0&0&1 \end{bmatrix}$$
and $v$ is:
$$\begin{bmatrix}
v_{1} \\
v_{2} \\
v_{3} \end{bmatrix}$$
When I plug in I get the matrix:
$$
\begin{bmatrix}3&-2&3 \\0&-2&3 \\-1&2&-3 \end{bmatrix}
\begin{bmatrix}v_{1} \\v_{2} \\v_{3} \end{bmatrix}
=
\begin{bmatrix}0 \\0 \\0 \end{bmatrix}
$$
I'm not so sure what to do after this.
 A: The characteristic polynomial is given by $|A - \lambda I| = 0$, hence:
$$-\lambda ^3+\lambda ^2+13 \lambda -13 = -(\lambda -1) \left(\lambda ^2-13\right) = 0$$
This shows one eigenvalue is $\lambda_1 = 1$.
To find the eigenvector, we solve $[A-\lambda_i I]v_i = [A - I]v_1=0$.
We perform Row-Reduced-Echelon-Form (RREF), yielding:
$$\begin{bmatrix}
 1 & 0 & 0 \\
 0 & 1 & -\frac{3}{2} \\
 0 & 0 & 0 \\
\end{bmatrix}v_1 = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$
Choose the eigenvector as $a = 0, c = 2 , b = 3$, so
$$ v_1 = \begin{bmatrix} 0 \\ 3 \\ 2\end{bmatrix}$$
A: if $1$ is an eigenvalue of $A,$ then you want to solve $$(A - I)x = 0 $$ not $(A + I)x = 0 $ as you are trying to do. when row reduce $A - I,$ i get $$\pmatrix{1&0&0\\0&1&-1.5\\0&0&0}$$ then an eigenvector corresponding to eignevalue $1$ is $$\pmatrix{0\\3\\2}$$
A: It looks like you've made a few small errors. To prove that $\lambda=1$ is an eigenvalue of 
$$
A=\begin{bmatrix}4&-2&3\\ 0&-1&3\\-1&2&-2\end{bmatrix}
$$
we must show that $\det(A-\lambda I)=0$. But
$$
A-I =
\begin{bmatrix}3&-2&3\\ 0&-2&3\\-1&2&-3\end{bmatrix}
$$
so 
$$
\det(A-I)
=3\begin{vmatrix}-2&3\\2&-3\end{vmatrix}-\begin{vmatrix}-2&3\\-2&3\end{vmatrix}
=3\cdot0-0=0
$$
This proves $\lambda=1$ is an eigenvalue of $A$.
Now, to find an eigenvector of $A$ corresponding to the eigenvalue $\lambda=1$ we must compute $\DeclareMathOperator{Null}{Null}\Null(A-I)$. But 
$$
\DeclareMathOperator{rref}{rref}\rref(A-I)=
\begin{bmatrix}1&0&0\\ 0&1&-\frac{2}{3}\\ 0&0&0\end{bmatrix}
$$
so $v=(0,1,3/2)$ is an eigenvector.
