Question

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

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

Short answer

For \(\lambda = 1\): \(\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\).

For \(\lambda = 2\): \(\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\2\end{array}} \right)\).

For \(\lambda = 3\): \(\left( {\begin{array}{*{20}{c}}{ - 1}\\1\\1\end{array}} \right)\).

Step-by-step solution

Definitions

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 \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

Find a basis of eigenspace for \(\lambda  = 1\)

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

As, \(\lambda = 1,2,3\) are the eigenvalue of the matrix \(A\), so they satisfy the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

For \(\lambda = 1\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

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

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

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

The obtained matrix is hard to reduce further. Write a system of equations corresponding to the obtained matrix.

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

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

\(\begin{array}{c}{x_1} = 0\\{x_2} = 1\\{x_3} = 0\end{array}\)

So, the general solution is given as:

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\\ = {x_2}{e_2}\end{array}\)

So, \({e_2} = \left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\) is the basis for the eigenspace for \(\lambda = 1\).

Find a basis of eigenspace for \(\lambda  = 2\)

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

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

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

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

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

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

Write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}{x_1} + \frac{1}{2}{x_3} = 0\\{x_2} - {x_3} = 0\\{x_3},{\rm{ free variable}}\end{array}\)

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

\(\begin{array}{c}{x_1} = - \frac{1}{2}\\{x_2} = 1\\{x_3} = 1\end{array}\)

So, the general solution is given as:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}\\1\\1\end{array}} \right)\)

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

Find a basis of eigenspace for \(\lambda  = 3\)

For \(\lambda = 3\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

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

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

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

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

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

Write a system of equations corresponding to the obtained matrix.

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

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

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

So, the general solution is given as:

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

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