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"?

There are many ways of finding eigenvectors and eigenvalues, but they all come down to solving a linear system of equations.  For example, in this case you can use the fact that $\lambda_1=6$ is an eigenvalue (which you can easily find by solving the characteristic equation $-\lambda^3 + 6\lambda^2 - 9\lambda = 0$), to solve for the eigenvector corresponding to that eigenvalue.  The eigenvector is the solution to the system of equations
$$
\begin{bmatrix}
-6 & 2 & 3 \\
1 & -6 & 3 \\
1 & 2 & -6
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}.
$$
Note that the coefficient matrix of this system is the identity matrix, except for the $-6$ in each diagonal entry, which is exactly what you would expect to see for an eigenvalue of 6.
This linear system is actually very easy to solve, even without using software, because the equations are all multiples of the first equation:
$$
\begin{align*}
-6x + 2y + 3z &= 0 \\
-6x + 2y + 3z &= 0 \\
-6x + 2y + 3z &= 0.
\end{align*}
$$
That means that the solution can only have $x=y=z$, so we can simplify to
$$
\begin{align*}
-6x + 2x + 3x &= 0 \\
-6x + 2x + 3x &= 0 \\
-6x + 2x + 3x &= 0,
\end{align*}
$$
which is just
$$
\begin{align*}
x &= 0 \\
x &= 0 \\
x &= 0.
\end{align*}
$$
Thus, the only solution to the linear system is the trivial solution $x=y=z=0$, which means that the only eigenvector corresponding to $\lambda_1=6$ is the trivial vector $(0,0,0)^T$.
Now that we have found one eigenvector, we can use it to find the other eigenvector.  We know that $\lambda_2=0$ is an eigenvalue (which you can easily find by solving the characteristic equation), so we can solve for the eigenvector corresponding to that eigenvalue.  The eigenvector is the solution to the system of equations
$$
\begin{bmatrix}
0 & 2 & 3 \\
1 & 0 & 3 \\
1 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}.
$$
Again, the linear system is very easy to solve.  The first equation implies that $z=-\frac{2}{3}y$, so we can substitute that value of $z$ into the second and third equations to get
$$
\begin{align*}
2y &= 3x \\
-\frac{4}{3}y &= 3x,
\end{align*}
$$
which is
$$
3x = \frac{4}{3}y.
$$
That means that we can choose $x$ and $y$ arbitrarily, as long as $x=\frac{4}{9}y$.  So the solution to the linear system is
$$
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
\frac{4}{9}y \\ y \\ -\frac{2}{3}y
\end{bmatrix}
=
y
\begin{bmatrix}
\frac{4}{9} \\ 1 \\ -\frac{2}{3}
\end{bmatrix}
=
y\begin{bmatrix}
\frac{4}{9} \\ 1 \\ -\frac{2}{3}
\end{bmatrix}.
$$
Note that this is a non-trivial eigenvector, which means that $\lambda_2=0$ is a non-zero eigenvalue.
Thus, the eigenvectors of the matrix are $\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}$ and $\begin{bmatrix}
\frac{4}{9} \\ 1 \\ -\frac{2}{3}
\end{bmatrix}$, and the eigenvalues are $\lambda_1=6$ and $\lambda_2=0$.
