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"?
 A: 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.
A: 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 $6$.
A: 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.
A: 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 $\lambda_2=0$.
