In Exercises 9–16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)

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 be 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 \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), where \(\lambda = 4\).

As, \(\lambda = 4\) is the eigenvalue of the matrix \(A\), so it satisfies the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

For \(\lambda = 4\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

\(\begin{array}{c}\left( {A - 4I} \right) = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right) - 4\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}4&0\\0&4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}6&{ - 9}\\4&{ - 6}\end{array}} \right)\end{array}\)

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

\(\left( {\begin{array}{*{20}{c}}6&{ - 9}&0\\4&{ - 6}&0\end{array}} \right)\)

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

\(\begin{gathered} \hfill \left( {\begin{array}{*{20}{c}} 6&{ - 9}&0 \\ 4&{ - 6}&0 \end{array}} \right)\xrightarrow{{{R_1} \to \frac{{{R_1}}}{6}}}\left( {\begin{array}{*{20}{c}} 1&{ - \frac{3}{2}}&0 \\ 4&{ - 6}&0 \end{array}} \right) \\ \hfill \xrightarrow{{{R_2} \to {R_2} - 4{R_1}}}\left( {\begin{array}{*{20}{c}} 1&{ - \frac{3}{2}}&0 \\ 0&0&0 \end{array}} \right) \\ \end{gathered} \)

Write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}{x_1} - \frac{3}{2}{x_2} = 0\\{x_2},{\rm{ free variable}}\end{array}\)

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

\(\begin{array}{c}{x_1} = \frac{3}{2}\\{x_2} = 1\end{array}\)

So, the general solution is given as:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\end{array}} \right)\)

So, \(\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\end{array}} \right)\) is the basis for the eigenspace for \(\lambda = 4\).