Finding Eigenvectors for $3 \times 3$ matrix with rows of zeros. For a $3 \times 3$ matrix:
$
   $[A]$  =   \begin{bmatrix}
        6 & -2  & 2 \\
        -2 & 3  & -1 \\
        2 & -1 & 3
        \end{bmatrix}
$
I have the eigenvalues: $\lambda = 2, 2, 8$
Now for each value I need to find eigenvectors:
Now, when solving for when $\lambda = 2$ and after applying the row operations, I am left with:
$$
\begin{bmatrix}
  4 & -2 & 2  \\   
  0 & 0 & 0  \\
  0 & 0 & 0 \\
  \end{bmatrix} \begin{bmatrix}
   x_1 \\ x_2 \\ x_3
   \end{bmatrix}
= \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}
$$
Now, they let $x_3 = k_1$ and $x_2 = k_2$
So, that: $$x_1 = \frac{k_2}{2} - \frac{k_1}{2}$$
Hence, $$X =[x_1, x_2, x_3] = [\frac{k_2}{2} - \frac{k_1}{2}, k_2, k_1]$$
That's okay. But what I don't understand is the following step:
$$ = k_1 [ \frac{-1}{2}, 0, 1 ] + k_2 [\frac{1}{2}, 1, 0]$$
And when we pick $k_1 = k_2 = 1$, we get the eigen vectors.
Is that last step done by taking something as common factor or factorizing the matrix values? Exactly how?
 A: The formula
$$X=k_1[-1/2,0,1]+k_2[1/2,1,0]$$
tells you, that for each cobination of $k_1,k_2$, the resulting vector satisfies $Ax=\lambda_1x$ where $\lambda_1=2$. Therefore the eigenvectors for $\lambda_1$ are $[-1/2,0,1]$ and $[1/2,1,0]$ 
The previous step 
$$[k_2/2-k_1/2,k_2,k_1]= k_1[-1/2,0,1]+k_2[1/2,1,0]$$
is basically a factorization. Alternatively, you can set $k_1=1,k_2=0$ to get the first eigenvector and $k_1=0,\ k_2= 1$ to get the second eigenvector. 
A: Recall that the set of all eigenvectors of matrix $A$ corresponding to eigenvalue $\lambda=2$ 
$$ V_2=\{ [\frac{k_2}{2} - \frac{k_1}{2}, k_2, k_1] \mid k_1,k_2 \in \mathbb R\}$$
forms a vector subspace of $\mathbb R^3$. Since every element of $V_2$ can be expressed as a linear combination 
$$ [\frac{k_2}{2} - \frac{k_1}{2}, k_2, k_1] =  k_1 [-\frac{1}{2},0,1] + k_2 [\frac{1}{2},1,0] \tag{1}$$
of two linearly independent vectors $[-\frac{1}{2},0,1]$ and $[\frac{1}{2},1,0]$, these two vectors form a basis of $V_2$ and we can describe $V_2$ as a vector space spanned by this basis.
Again, for every eigenvalue, there is not one, but infinitely many eigenvectors that correspond to this eigenvalue, and they all form a vector space. One way to describe this vector space is to find one of its basis as I did above.
