91Ó°ÊÓ

Eigenvalue: Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to an eigenvalue of the matrix \(A\) if there exist a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

Eigenvector: For a \(n \times n\) matrix \(A\), whose eigenvalue is \(\lambda \), the set of a subspace of \({\mathbb{R}^n}\) is known as an eigenspace, where the set of the subspace of is the set of all the solutions of \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

The given matrix is \(A = \left( {\begin{array}{*{20}{c}}5&0\\2&1\end{array}} \right)\), where \(\lambda = 1,5\).

As, \(\lambda = 1,5\) are the eigenvalue of the matrix \(A\), so they must satisfy the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

For \(\lambda = 1\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

\(\begin{array}{c}\left( {A - 1I} \right) = \left( {\begin{array}{*{20}{c}}5&0\\2&1\end{array}} \right) - 1\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4&0\\2&0\end{array}} \right)\end{array}\)

Write the obtained matrix in the form of an augmented matrix, where for \(A{\bf{x}} = 0\), the augmented matrix given by \(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right)\).

\(\left( {\begin{array}{*{20}{c}}4&0&0\\2&0&0\end{array}} \right)\)

The obtained matrix cannot be reduced further, so write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}4{x_1} = 0\\2{x_1} = 0\\{x_2},{\rm{ free variable}}\end{array}\)

As \({x_2}\) is a free variable, let \({x_2} = 1\). Then,

\(\begin{array}{c}{x_1} = 0\\{x_2} = 1\end{array}\)

So, the general solution is given as:

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\\ = {x_2}{e_2}\end{array}\)

As, \({e_2} = \left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\). So \({e_2} = \left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\) is the basis for the eigenspace for \(\lambda = 1\).

For \(\lambda = 5\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

\(\begin{array}{c}\left( {A - 5I} \right) = \left( {\begin{array}{*{20}{c}}5&0\\2&1\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0&0\\2&{ - 4}\end{array}} \right)\end{array}\)

Write the obtained matrix in the form of an augmented matrix, where for \(A{\bf{x}} = 0\), the augmented matrix given by \(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right)\).

\(\left( {\begin{array}{*{20}{c}}0&0&0\\2&{ - 4}&0\end{array}} \right)\)

The obtained matrix cannot be reduced further, so write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}2{x_1} - 4{x_2} = 0\\{x_1} = 2{x_2}\end{array}\)

So, \({x_2}\) is a free variable, let \({x_2} = 1\). Then,

\(\begin{array}{c}{x_1} = 2\left( 1 \right)\\ = 2\end{array}\)

So, the general solution is given as:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\)

So, \(\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\) is the basis for the eigenspace for \(\lambda = 5\).