Confusion determining eigenvalues Matrix $A$:
$$A = \begin{bmatrix}
    -5 &14& -8 \\
    -9 &16& -6 \\
    -9 &11& -1
\end{bmatrix} $$
Determining eigenvalue of Matrix $A - 1I$ hence $\lambda = 1$ (other eigenvalues are 4 and 5).
Matrix $A - 1I:$ 
$$\begin{bmatrix}
    -6 &14& -8 \\
    -9 &15& -6 \\
    -9 &11& -2
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}$$
1. Figuring out each $x_i$ so the above equation complies will give me the eigenvector for eigenvalue $\lambda = 1$ right? 
2. the sum of all $x_i$ must be $0$? 
3. How would you go about determining $x_i$, do you just play around till you get a right combination? 
 A: Using gaussian elimination, your system is equivalent to the following, writing only the matrix, not the $x_i$ nor the right hand side since there are only zeros in the RHS anyway:
$$\left(\begin{matrix}
-6 &14& -8 \\
-9 &15& -6 \\
-9 &11& -2
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
-9 &15& -6 \\
-9 &11& -2
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
0 &-6& 6 \\
0 &-10& 10
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
0 &-6& 6 \\
0 &0& 0
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
0 &-6& 6 \\
0 &0& 0
\end{matrix}\right)
\left(\begin{matrix}
x_1 \\ x_2 \\ x_3
\end{matrix}\right)
=\left(\begin{matrix}
0 \\ 0 \\ 0
\end{matrix}\right)$$
That is, $x_3$ is arbitrary, then $x_2=x_3$ from the second equation, and finally $x_1=-\frac13(4x_3-7x_2)=x_3$.
Therefore, $(1,1,1)^T$ is an eigenvector of $A$ for the eigenvalue $1$.
When solving such system, if the eigenvalue is correct, you know for sure there is at least one arbitrary component. There are more if the eigenspace has dimension greater than $1$.
