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

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

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

As, \(\lambda = 4\) are the eigenvalue of the matrix \(A\), so they satisfy the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

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

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

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

\(\left( {\begin{array}{*{20}{c}}{ - 1}&0&2&0&0\\1&{ - 1}&1&0&0\\0&1&{ - 3}&0&0\\0&0&0&0&0\end{array}} \right)\)

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

\(\begin{gathered} \hfill \left( {\begin{array}{*{20}{c}} { - 1}&0&2&0&0 \\ 1&{ - 1}&1&0&0 \\ 0&1&{ - 3}&0&0 \\ 0&0&0&0&0 \end{array}} \right)\xrightarrow{{{R_1} \to - {R_1}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - 2}&0&0 \\ 1&{ - 1}&1&0&0 \\ 0&1&{ - 3}&0&0 \\ 0&0&0&0&0 \end{array}} \right) \\ \hfill \xrightarrow({}){{{R_2} \to {R_2} - {R_1}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - 2}&0&0 \\ 0&{ - 1}&3&0&0 \\ 0&1&{ - 3}&0&0 \\ 0&0&0&0&0 \end{array}} \right) \\ \hfill \xrightarrow({}){{{R_2} \to - {R_2}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - 2}&0&0 \\ 0&1&{ - 3}&0&0 \\ 0&1&{ - 3}&0&0 \\ 0&0&0&0&0 \end{array}} \right) \\ \hfill \xrightarrow[{}]{{{R_3} \to {R_3} - {R_2}}}\left( {\begin{array}{*{20}{c}} 1&0&{ - 2}&0&0 \\ 0&1&{ - 3}&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{array}} \right) \\ \end{gathered} \)

Write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}{x_1} - 2{x_3} = 0\\{x_2} - 3{x_3} = 0\\{x_3},{x_4},{\rm{ free variables}}\end{array}\)

As \({x_3},{x_4}\) are free variables, then,

\(\begin{array}{l}{x_1} = 2{x_3}\\{x_2} = 3{x_3}\end{array}\)

So, the general solution is given as:

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2{x_3}}\\{3{x_3}}\\{{x_3}}\\{{x_4}}\end{array}} \right)\\ = {x_3}\left( {\begin{array}{*{20}{c}}2\\3\\1\\0\end{array}} \right) + {x_4}\left( {\begin{array}{*{20}{c}}0\\0\\0\\1\end{array}} \right)\end{array}\)

So, \(\left\{ {\left( {\begin{array}{*{20}{c}}2\\3\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}0\\0\\0\\1\end{array}} \right)} \right\}\) is the basis for the eigenspace for \(\lambda = 4\).