Question

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)

Answer 1

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

{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.