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"?

It's easy to see that $0$ is a multiple eigenvalue with multiplicity two as we can exhibit two linearly independent vector of the kernel (since all columns are obviously a multiple of the first one). Then the last eigenvalue $\lambda$ can be found using the Trace as the sum of the diagonal elements is also equal to the sum of the eigenvalues counted with their multiplicity : $0 + 0 + \lambda = 1+2+3$. So the eigenvalues are $0$ and $6$.
