How do you find the Eigenvalue and Eigenvectors of a matrix?
The matrix is \[\left( {\begin{array}{*{20}{c}}
  0&4&0 \\
  { - 1}&{ - 4}&0 \\
  0&0&{ - 2}
\end{array}} \right)\]

Hint: Set up the characteristic equation, using $|A - \lambda I| = 0$
Solve the characteristic equation, giving us the eigenvalue
Substitute the eigenvalue into the two equations given by $|A - \lambda I|$
Choose a convenient value for ${x_1}$, then find ${x_2}$
The resulting values form the corresponding eigenvectors of a given matrix.

Complete step-by-step solution:
Let us consider the given matrix \[\left( {\begin{array}{*{20}{c}}
  0&4&0 \\
  { - 1}&{ - 4}&0 \\
  0&0&{ - 2}
\end{array}} \right) = A\]
If non-zero $e$ is an eigenvector of the $3$ by $3$ matrix $A$ , then $Ae = \lambda e$ For some scalar $\lambda $.
This scalar is called an eigenvalue of $A$
This may be rewritten as
$ \Rightarrow Ae = \lambda Ie$
And inturn we write this as
$ \Rightarrow \left( {A - \lambda I} \right)e = 0$
Therefore the characteristic equation is
$ \Rightarrow A - \lambda I = 0$
Now substitute the matrix value, we get
\[ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {0 - \lambda }&4&0 \\
  { - 1}&{ - 4 - \lambda }&0 \\
  0&0&{ - 2 - \lambda }
\end{array}} \right) = 0\]
Now we can expand the determinant
$ \Rightarrow \left( {0 - \lambda } \right)\left( {\left( { - 4 - \lambda } \right)\left( { - 2 - \lambda } \right) - 0} \right) - 4\left( {\left( { - 1} \right)\left( { - 2 - \lambda } \right) - 0} \right) + 0\left( {0 - 0} \right) = 0$
Reduce the equation by multiplying inside brackets, we get
$ \Rightarrow - \lambda \left( {\left( { - 4 - \lambda } \right)\left( { - 2 - \lambda } \right)} \right) - 4\left( {2 + \lambda } \right) = 0$
On simplify the term and we get
$ \Rightarrow - \lambda \left( {8 + 4\lambda + 2\lambda + {\lambda ^2}} \right) - 8 - 4\lambda = 0$
Let us multiply we get,
$ \Rightarrow - 8\lambda - 4{\lambda ^2} - 2{\lambda ^2} - {\lambda ^3} - 8 - 4\lambda = 0$
On cancel the term and we get
$ \Rightarrow - {\lambda ^3} - 6{\lambda ^2} - 12\lambda - 8 = 0$
Taking minus common, we get
$ \Rightarrow {\lambda ^3} + 6{\lambda ^2} + 12\lambda + 8 = 0$
Now by factorizing the above equation
$ \Rightarrow {\left( {\lambda + 2} \right)^3} = 0$
Therefore we get
$ \Rightarrow \lambda = - 2$
The eigenvalue is $ - 2$
Now substitute these Eigen value in the characteristic equation, we get
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {0 - \lambda }&4&0 \\
  { - 1}&{ - 4 - \lambda }&0 \\
  0&0&{ - \lambda - 2}
\end{array}} \right)$
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  { - \left( { - 2} \right)}&4&0 \\
  { - 1}&{ - \left( { - 2 - 4} \right)}&0 \\
  0&0&{ - \left( { - 2 - 2} \right)}
\end{array}} \right)$
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  2&4&0 \\
  { - 1}&{ - 2}&0 \\
  0&0&0
\end{array}} \right)$
Perform row operations, we get
Multiply second row with $2$
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  2&4&0 \\
  { - 2}&{ - 4}&0 \\
  0&0&0
\end{array}} \right)$
Now subtract second row from first row, we get
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  2&4&0 \\
  0&0&0 \\
  0&0&0
\end{array}} \right)$
Now divide first row by $2$ we get
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  0&0&0 \\
  0&0&0
\end{array}} \right)$
Now solve the matrix equation by
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  0&0&0 \\
  0&0&0
\end{array}} \right)\,\left( {\begin{array}{*{20}{c}}
  {{x_1}} \\
  {{x_2}} \\
  {{x_3}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  0
\end{array}} \right)$
From the above matrix equation we can find
$ \Rightarrow {x_1} + 2{x_2} = 0$
By considering ${x_2} = {x_2}$ and ${x_3} = {x_3}$
We can now find the value of ${x_1}$ we get
$ \Rightarrow {x_1} = - 2{x_2}$
$ \Rightarrow x = \left( {\begin{array}{*{20}{c}}
  { - 2{x_1}} \\
  {{x_2}} \\
  {{x_3}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 2} \\
  1 \\
  0
\end{array}} \right){x_2} + \left( {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  1
\end{array}} \right){x_3}$
$ \Rightarrow x = \left( {\begin{array}{*{20}{c}}
  { - 2} \\
  1 \\
  0
\end{array}} \right)$

Note: Here, we were dealing with a $3 \times 3$ system, and we found $3$ eigenvalue and $3$ corresponding eigenvectors.
If we had a $2 \times 2$ system, we would have found $2$ eigenvalue and $2$ corresponding eigenvectors.
In general, $n \times n$ system will produce $n$ eigenvalue and $n$ corresponding eigenvectors.
We could have easily chosen same value for ${x_{1\,}}$ and ${x_2}$, however it's usually more meaningful to choose a convenient starting value(usually for ${x_1}$ ) and then derive the resulting remaining values.