Question: Is \(\lambda = 3\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}1&2&2\\3&{ - 2}&1\\0&1&1\end{array}} \right)\)? 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 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 \(A\).

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

According to the definition of eigenvalue, \(\lambda = 3\) is the eigenvalue of the matrix \(A\), if satisfies the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

Substitute \(\lambda = 3\) into \(A{\bf{x}} = \lambda {\bf{x}}\).

\(A{\bf{x}} = 3{\bf{x}}\)

The obtained equation is equivalent to \(\left( {A - 3I} \right){\bf{x}} = 0\), which is a homogeneous equation.

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

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

It can be said that \(\lambda = 3\) will be the eigenvalue of the given matrix if \(\left( {A - 3I} \right)\) 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 \(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right)\).

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

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

\(\begin{gathered} \hfill \left( {\begin{array}{*{20}{c}} { - 2}&2&2&0 \\ 3&{ - 5}&1&0 \\ 0&1&{ - 2}&0 \end{array}} \right)\underrightarrow {{R_1} \to - \frac{{{R_1}}}{2}}\left( {\begin{array}{*{20}{c}} 1&{ - 1}&{ - 1}&0 \\ 3&{ - 5}&1&0 \\ 0&1&{ - 2}&0 \end{array}} \right) \\ \hfill \underrightarrow {{R_2} \to {R_2} -3{R_1}}\left( {\begin{array}{*{20}{c}} 1&{ - 1}&{ - 1}&0 \\ 0&{ - 2}&4&0 \\ 0&1&{ -2}&0 \end{array}} \right) \\ \hfill \underrightarrow {{R_2} \to - \frac{{{R_2}}}{2}}\left( {\begin{array}{*{20}{c}} 1&{ - 1}&{ - 1}&0 \\ 0&1&{ - 2}&0 \\ 0&1&{ - 2}&0 \end{array}} \right) \\ \hfill \underrightarrow {{R_1} \to {R_1} + {R_2}}\left( {\begin{array}{*{20}{c}} 1&0&{ - 3}&0 \\ 0&1&{ - 2}&0 \\ 0&1&{ - 2}&0 \end{array}} \right) \\ \hfill \underrightarrow {{R_3} \to {R_3} - {R_2}}\left( {\begin{array}{*{20}{c}} 1&0&{ - 3}&0 \\ 0&1&{ - 2}&0 \\ 0&0&0&0 \end{array}} \right) \\ \end{gathered} \)

From the obtained matrix, it can be seen that \(\left( {A - 3I} \right){\bf{x}} = 0\) has a nontrivial solution, which does not need to find, but according to the definition of eigenvalue, if \(\left( {A - \lambda I} \right){\bf{x}} = 0\) has the nontrivial solution, then the scaler \(\lambda \) is the eigenvalue of the matrix \(A\). So \(\lambda = 3\) is the eigenvalue of the given matrix.

Write the obtained matrix in the form of equations.

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

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

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

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.

\({\bf{x}} = \left( {\begin{array}{*{20}{c}}3\\2\\1\end{array}} \right)\)

Hence, one of the corresponding eigenvectors is \({\bf{x}} = \left( {\begin{array}{*{20}{c}}3\\2\\1\end{array}} \right)\).