<s>[INST] 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"?
 [/INST] One way to instantly see that $0$ is an eigenvalue is that the three columns are clearly not linearly independent --- in fact, they're all constant multiples of each other! Next, any dependency vector of those columns gives you an eigenvector for the eigenvalue $0$. For example, clearly the expression
$$\text{column 1 + column 2 - column 3}
$$
evaluates to the zero column vector. Putting the three coefficients $1, 1, -1$ of that expression into a column vector you therefore get the following eigenvector for the eigenvalue $0$ namely $$ \begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}$$
I'm sure, without looking back at your own post, you can easily "see" another dependency vector to get another eigenvector for the eigenvalue $0$ which is linearly independent of the one just given.
For the remaining eigenvector, you could notice that every column has constant coefficients, and therefore the expression
$$\text{column 1 + column 2 + column 3}
$$
evaluates to a column vector that also has constant coefficients. Since the three coefficients $1, 1, 1$ of that expression are constant, putting those three coefficients into column vector therefore gives you an eigenvector
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
And since the actual sum of the columns is
\begin{pmatrix}
6 \\
6 \\
6
\end{pmatrix}
you immediately conclude that $6$ is the eigenvalue.
</s>