Why is this an eigenvector of this matrix? The matrix is 
$$
\begin{bmatrix}
1 & 1 & -3 \\
2 & 0 & 6 \\
1 & -1 & 5
\end{bmatrix}
$$
I understand where the eigenvector $[-3, 0, 1]$ comes from because the eigenvalue is $\lambda = 2$ and when you substitute it into the matrix and then row reduce you get
$$
\begin{bmatrix}
1 & 0 & 3 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
But the answer also puts $[1, 1, 0]$ as an eigenvector. Any idea where this comes from?
 A: $$\det\Bigg(\begin{bmatrix}
   1-\lambda       & 1 & -3 \\
    2       & -\lambda & 6 \\
   1       & -1 & 5-\lambda
\end{bmatrix}\Bigg)=(1-\lambda)(-\lambda(5-\lambda)+6)-(2(5-\lambda)-6)-3(-2+\lambda)$$
$$\det(A-\lambda I) = (\lambda-2)^3=0$$
$$A-2 I =\begin{bmatrix}
   -1       & 1 & -3 \\
    2       & -2 & 6 \\
   1       & -1 & 3
\end{bmatrix}$$
From here you can see that the bottom two rows are just multiples of the first row. So
$$\begin{bmatrix}
   -1       & 1 & -3 \\
    2       & -2 & 6 \\
   1       & -1 & 3
\end{bmatrix}\begin{bmatrix}
   x        \\
    y       \\
   z      
\end{bmatrix}=\begin{bmatrix}
   -1       & 1 & -3 \\
   0       & 0 & 0 \\
   0       & 0 & 0
\end{bmatrix}\begin{bmatrix}
   x        \\
    y       \\
   z      
\end{bmatrix}=0$$
This means that $$x=y-3z$$ So we get the eigenvector is
$$\begin{bmatrix}
   y-3z        \\
    y       \\
   z      
\end{bmatrix}=\begin{bmatrix}
   y        \\
    y       \\
   0      
\end{bmatrix}+\begin{bmatrix}
   -3z        \\
    0       \\
   z      
\end{bmatrix}= y\begin{bmatrix}
   1        \\
    1       \\
   0      
\end{bmatrix}+z\begin{bmatrix}
   -3        \\
    0       \\
   1     
\end{bmatrix}$$
Which are your two linearly independant eigenvectors.
A: If the given matrix is multiplied by the transpose of $(1,1,0)$ the result is the transpose of $(2,2,0)$ which makes $(1,1,0)$ an eigenvector with eigenvalue $2.$
A: Let me give you a new way to calculate eigenvectors that you can use to solve this problem. When $v$ is an eiganvector, $v$ satisfies $Av=\lambda v$ for some $\lambda\neq0$. Thus $Av-\lambda v = 0$ where $0$ is the zero matrix, so $(A-\lambda I)v = 0$. This tells us that $\det(A-\lambda I)=0$, so try computing this determinant (with $\lambda$ a variable) and solving the resultant cubic to obtain all the eiganvalues. Then, for each $\lambda_i$, solve $Av=\lambda_i v$.
