91Ó°ÊÓ

If is an\(n \times n\)matrix, then \(det\left( {A - \lambda I} \right) = 0\), is called the characteristic equation of a matrix \(A\).

It is given that \(A = \left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&{\,\,3}\end{aligned}} \right)\) and \(I = \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\) is the identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\) as shown below:

\(\begin{aligned}{}A - \lambda I &= \left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&{\,\,3}\end{aligned}} \right) - \lambda \left( {\begin{aligned}{}1&0\\0&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{1 - \lambda }&{}&{ - 1}\\1&{}&{3 - \lambda }\end{aligned}} \right)\end{aligned}\)

Now calculate the determinant of the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{}{1 - \lambda }&{}&{ - 2}\\1&{}&{3 - \lambda }\end{aligned}} \right)\\ &= \left( {1 - \lambda } \right)\left( {3 - \lambda } \right) + 2\\ &= {\lambda ^2} - 4\lambda + 5\end{aligned}\)

So, the characteristic equation of the matrix \(A\) is \({\lambda ^2} - 4\lambda + 5 = 0\).

Ifis an\(n \times n\)matrix, then\(det\left( {A - \lambda I} \right) = 0\), is called the characteristic equation of a matrix\(A\).

It is given that\(A = \left( {\begin{aligned}{}1&{ - 2}\\1&{\,\,3}\end{aligned}} \right)\)and\(I = \left( {\begin{aligned}{}1&0\\0&1\end{aligned}} \right)\)is the identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}A - \lambda I &= \left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&{\,\,3}\end{aligned}} \right) - \lambda \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{1 - \lambda }&{}&{ - 1}\\1&{}&{3 - \lambda }\end{aligned}} \right)\end{aligned}\)

Now calculate the determinant of the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{}{1 - \lambda }&{}&{ - 2}\\1&{}&{3 - \lambda }\end{aligned}} \right)\\ &= \left( {1 - \lambda } \right)\left( {3 - \lambda } \right) + 2\\ &= {\lambda ^2} - 4\lambda + 5\end{aligned}\)

So, the characteristic equation of the matrix \(A\) is \({\lambda ^2} - 4\lambda + 5 = 0\).

Thus, the eigenvalues of\(A\)are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, solve the characteristic equation\({\lambda ^2} - 4\lambda + 5 = 0\), as follows:

For the quadratic equation, \(a{x^2} + bx + c = 0\) , the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\) .

Thus, the solution of the characteristic equation \({\lambda ^2} - 4\lambda + 5 = 0\) is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - 4\lambda + 5 &= 0\\\lambda &= \frac{{ - \left( { - 4} \right) \pm \sqrt {{4^2} - 4\left( 5 \right)} }}{2}\\ &= \frac{{4 \pm \sqrt { - 4} }}{2}\\ &= 2 \pm i\end{aligned}\)

The eigenvalues of \(A\) are \(\lambda = 2 + i\) and \(\lambda = 2 - i\) .

For the Eigenvalue, \({\lambda _i}\), matrix A satisfies the system of equations \(\left( {A - {\lambda _i}} \right){\rm{x}} = {\rm{0}}\).

For the Eigenvector\(\lambda = 2 + i\), the system \(\left( {A - \lambda I} \right)\)is solved as follows:

\(\begin{aligned}{}A - \left( {2 + i} \right)I &= \left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&{\,\,3}\end{aligned}} \right) - \left( {2 + i} \right)\left( {\begin{aligned}{}1&0\\{\,0}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 1 - i}&{}&{ - 2}\\1&{}&{1 - i}\end{aligned}} \right)\end{aligned}\)

The system of equations \(\left( {A - \lambda I} \right)x = 0\)gives the following equations:

\(\begin{aligned}{}\left( {\begin{aligned}{}{ - 1 - i}&{}&{ - 2}\\1&{}&{1 - i}\end{aligned}} \right) = 0\\\left( { - 1 - i} \right){x_1} - 2{x_2} = 0\\{x_1} + \left( {1 - i} \right){x_2} = 0\end{aligned}\)

Both the equations are equivalent to the equation, \({x_1} = - \left( {1 - i} \right){x_2}\), in which \({x_2}\) the variable is free. So, the general solution is \({x_2}\left( \begin{aligned}{} - 1 + i\\\,\,\,\,\,\,1\end{aligned} \right)\) and the corresponding Eigenvector is \({{\bf{v}}_1} = \left( \begin{aligned}{} - 1 + i\\\,\,\,\,\,\,1\end{aligned} \right)\).

For the Eigenvalue,\(\lambda = 2 - i\), the system\(\left( {A - \lambda I} \right)\)is solved as follows:

\(\begin{aligned}{c}A - \left( {2 - i} \right)I &= \left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&{\,\,3}\end{aligned}} \right) - \left( {2 - i} \right)\left( {\begin{aligned}{}1&{}&0\\{\,0}&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 1 + i}&{}&{ - 2}\\1&{}&{1 + i}\end{aligned}} \right)\end{aligned}\)

The system of equations \(\left( {A - \lambda I} \right)x = 0\)gives the following equations:

\(\begin{aligned}{c}\left( {\begin{aligned}{}{ - 1 + i}&{}&{ - 2}\\1&{}&{1 + i}\end{aligned}} \right) = 0\\\left( { - 1 + i} \right){x_1} - 2{x_2} = 0\\{x_1} + \left( {1 + i} \right){x_2} = 0\end{aligned}\)

Both the equations are equivalent to the equation, \({x_1} = - \left( {1 + i} \right){x_2}\), in which \({x_2}\) the variable is free. So, the general solution is \({x_2}\left( \begin{aligned}{} - 1 - i\\\,\,\,\,\,\,1\end{aligned} \right)\) and the corresponding Eigenvector is \({{\bf{v}}_2} = \left( \begin{aligned}{} - 1 - i\\\,\,\,\,\,\,1\end{aligned} \right)\).