Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick The matrix is: $ 
 \begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{pmatrix}
$
The solution says that
$ B\cdot
 \begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
= 
 \begin{pmatrix}
6 \\
6 \\
6\end{pmatrix}$
$ B\cdot
 \begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}
= 
 \begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
$ B\cdot
 \begin{pmatrix}
3 \\
0 \\
-1
\end{pmatrix}
= 
 \begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
Thus the eigenvalues are $ \lambda_{1}=6,\lambda_{2}=0  $
My question is, how can I easily find
$\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$ and $\begin{pmatrix} 6 \\
6 \\
6\end{pmatrix}$?
Is there any way to see it "quickly"?

The range of the given matrix is spanned by one vector:
$$
           \left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right]
$$
Therefore, any eigenvector (which must be non-zero by definition) must be a scalar multiple of this vector, or it must be in the null space of the given matrix. The above is an eigenvector of the given matrix. The null space is spanned by
$$
       \left[\begin{array}{r}2 \\ -1 \\  0\end{array}\right],\left[\begin{array}{r}3 \\ 0 \\ -1 \end{array}\right]
$$
You can easily rewrite this null space in terms of the vectors given in your statement of the problem.
