Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

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

The Diagonalization Theorem:An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

Consider the given diagonalization matrix of \(A\) as \(A = \left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}&0&{ - 1}\\0&1&2\\1&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&0&1\\2&1&4\\{ - 1}&0&{ - 2}\end{array}} \right)\).

Compare diagonalization matrix of \(A\) with \(A = PD{P^{ - 1}}\).

\[\begin{array}{c}P = \left( {\begin{array}{*{20}{c}}{ - 2}&0&{ - 1}\\0&1&2\\1&0&0\end{array}} \right)\\D = \left( {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&4\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}0&0&1\\2&1&4\\{ - 1}&0&{ - 2}\end{array}} \right)\end{array}\]

According to the diagonal entries of the matrix, \(D\) there are three eigenvalues.

\({\lambda _1} = 5\), \({\lambda _2} = 4\)and\({\lambda _3} = 5\)

The three eigenvectors are columns of the matrix \(P\).

\[{{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{ - 2}\\0\\1\end{array}} \right), {{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right), {{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\end{array}} \right)\]

Thus, the basis for eigenvalue \(\lambda = 5\) is \(\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{ - 2}\\0\\1\end{array}} \right)\) and the basis for eigenvalue \(\lambda = 4\) is \(\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\end{array}} \right)\).