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 exists 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 \(A = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\1&{ - 3}&0\\4&{ - 13}&1\end{array}} \right)\), where \(\lambda = - 2\).

As, \(\lambda = - 2\) are the eigenvalue of the matrix \(A\), so they satisfy the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

For \(\lambda = - 2\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

\(\begin{array}{c}\left( {A + 2I} \right) = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\1&{ - 3}&0\\4&{ - 13}&1\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\1&{ - 3}&0\\4&{ - 13}&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&0&{ - 1}\\1&{ - 1}&0\\4&{ - 13}&3\end{array}} \right)\end{array}\)

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

\(\left( {\begin{array}{*{20}{c}}3&0&{ - 1}&0\\1&{ - 1}&0&0\\4&{ - 13}&3&0\end{array}} \right)\)

The obtained matrix is not in a reduced form, so reduce it in row echelon form by applying row operations.

\(\begin{gathered} \hfill \left( {\begin{array}{*{20}{c}} 3&0&{ - 1}&0 \\ 1&{ - 1}&0&0 \\ 4&{ - 13}&3&0 \end{array}} \right)\xrightarrow{{{R_1} \to \frac{{{R_1}}}{3}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - \frac{1}{3}}&0 \\ 1&{ -1}&0&0 \\ 4&{ - 13}&3&0 \end{array}} \right) \\ \hfill \xrightarrow({{R_3} \to {R_3} - 4{R_1}}){{{R_2} \to {R_2} - {R_1}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - \frac{1}{3}}&0 \\ 0&{ - 1}&{\frac{1}{3}}&0 \\ 0&{ - 13}&{\frac{{13}}{3}}&0 \end{array}} \right) \\ \hfill \xrightarrow{{{R_2} \to - {R_2}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - \frac{1}{3}}&0 \\ 0&1&{ - \frac{1}{3}}&0 \\ 0&{ - 13}&{\frac{{13}}{3}}&0 \end{array}} \right) \\ \hfill \xrightarrow{{{R_3} \to {R_3} + 13{R_2}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - \frac{1}{3}}&0 \\ 0&1&{ - \frac{1}{3}}&0 \\ 0&0&0&0 \end{array}} \right) \\ \end{gathered} \)

Write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}{x_1} - \frac{1}{3}{x_3} = 0\\{x_2} - \frac{1}{3}{x_3} = 0\\{x_3},{\rm{ free variable}}\end{array}\)

As \({x_3}\) is a free variable, let \({x_3} = 1\). Then,

\(\begin{array}{c}{x_1} = \frac{1}{3}\\{x_2} = \frac{1}{3}\\{x_3} = 1\end{array}\)

So, the general solution is given as:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{1}{3}}\\1\end{array}} \right)\)

So, \(\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{1}{3}}\\1\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}}1\\1\\3\end{array}} \right)\) is the basis for the eigenspace for \(\lambda = - 2\).