Question: Is \(\lambda = 4\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}3&0&{ - 1}\\2&3&1\\{ - 3}&4&5\end{array}} \right)\)? If so, find one corresponding eigenvector.

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 \({\bf{x}}\) for \(A{\bf{x}} = \lambda {\bf{x}}\).

Denote the given matrix by \(A\).

\(A = \left( {\begin{array}{*{20}{c}}3&0&{ - 1}\\2&3&1\\{ - 3}&4&5\end{array}} \right)\)

According to the definition of eigenvalue, \(\lambda = 4\) is the eigenvalue of the matrix \(A\), if satisfies the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

Substitute \(\lambda = 4\) into \(A{\bf{x}} = \lambda {\bf{x}}\).

\(A{\bf{x}} = 4{\bf{x}}\)

The obtained equation equivalent to \(\left( {A - 4I} \right){\bf{x}} = 0\), which is a homogeneous equation.

So, first solve the matrix \(\left( {A - 4I} \right)\) by using \(A = \left( {\begin{array}{*{20}{c}}3&0&{ - 1}\\2&3&1\\{ - 3}&4&5\end{array}} \right)\).

\(\begin{array}{c}\left( {A - 4I} \right) = \left( {\begin{array}{*{20}{c}}3&0&{ - 1}\\2&3&1\\{ - 3}&4&5\end{array}} \right) - 4\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&0&{ - 1}\\2&3&1\\{ - 3}&4&5\end{array}} \right) - \left( {\begin{array}{*{20}{c}}4&0&0\\0&4&0\\0&0&4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}&0&{ - 1}\\2&{ - 1}&1\\{ - 3}&4&1\end{array}} \right){\rm{ - - - - - - - }}\left( 1 \right)\end{array}\)

It can be said that \(\lambda = 4\) will be the eigenvalue of the given matrix if \(\left( {A - 4I} \right)\) is invertible.

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

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

Now, reduce the obtained matrix into row echelon form by using the row operations.

\(\begin{gathered} \hfill \left( {\begin{array}{*{20}{c}} { - 1}&0&{ - 1}&0 \\ 2&{ - 1}&1&0 \\ { - 3}&4&1&0 \end{array}} \right)\underrightarrow {{R_1} \to - {R_1}}\left( {\begin{array}{*{20}{c}} 1&0&1&0 \\ 2&{ - 1}&1&0 \\ { - 3}&4&1&0 \end{array}} \right) \\ \hfill \underrightarrow {{R_2} \to {R_2} - 2{R_1}}\left( {\begin{array}{*{20}{c}} 1&0&1&0 \\ 0&{ - 1}&{ - 1}&0 \\ { - 3}&4&1&0 \end{array}} \right) \\ \hfill \underrightarrow {{R_3} \to {R_3} + 3{R_1}}\left( {\begin{array}{*{20}{c}} 1&0&1&0 \\ 0&{ - 1}&{ - 1}&0 \\ 0&4&4&0 \end{array}} \right) \\ \hfill \underrightarrow {{R_3} \to {R_3} + 4{R_2}}\left( {\begin{array}{*{20}{c}} 1&0&1&0 \\ 0&{ - 1}&{ - 1}&0 \\ 0&0&0&0 \end{array}} \right) \\ \end{gathered} \)

From the obtained matrix, it can be seen that \(\left( {A - 4I} \right){\bf{x}} = 0\) has a nontrivial solution, which does not need to find. Still, according to the definition of eigenvalue, if \(\left( {A - \lambda I} \right){\bf{x}} = 0\) has a nontrivial solution, then scaler \(\lambda \) is the eigenvalue of the matrix \(A\). So \(\lambda = 4\) is the eigenvalue of the given matrix.

Write the obtained matrix in the form of equations.

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

As \({x_3}\) is a free variable, let \({x_3} = 1\). Then find \({x_1}\), and \({x_2}\) by using \({x_3} = 1\).

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

The eigenvector is given by \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\). So, write the eigenvector by using the obtained values.

\({\bf{x}} = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\1\end{array}} \right)\)

Hence, one of the corresponding eigenvectors is \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\1\end{array}} \right)\).