Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

19. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{{\bf{ - 3}}}&{\bf{0}}&{\bf{9}}\\{\bf{0}}&{\bf{3}}&{\bf{1}}&{{\bf{ - 2}}}\\{\bf{0}}&{\bf{0}}&{\bf{2}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{2}}\end{array}} \right)\)

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 matrix \(A = \left( {\begin{array}{*{20}{c}}5&{ - 3}&0&9\\0&3&1&{ - 2}\\0&0&2&0\\0&0&0&2\end{array}} \right)\). Since it is given the eigenvalues of the matrix are \(5\), \(3\), \(2\) and \(2\).

Write the matrix form for finding the eigenvector.

\(\begin{array}{c}\left( {A - 5I} \right){\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}5&{ - 3}&0&9\\0&3&1&{ - 2}\\0&0&2&0\\0&0&0&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}5&0&0&0\\0&5&0&0\\0&0&5&0\\0&0&0&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}0&{ - 3}&0&9\\0&{ - 2}&1&{ - 2}\\0&0&{ - 3}&0\\0&0&0&{ - 3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

Therefore, the system of equations is shown below:

\(\begin{array}{c} - 3{x_2} + 9{x_4} = 0\\ - 2{x_2} + {x_3} - 2{x_4} = 0\\ - 3{x_3} = 0\\ - 3{x_4} = 0\end{array}\)

The general solution is,

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

Therefore, the eigenvector is \(\left\{ {{{\rm{v}}_1}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}1\\0\\0\\0\end{array}} \right)} \right\}\).

Write the matrix form for finding the eigenvector.

\(\begin{array}{c}\left( {A - 3I} \right){\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}5&{ - 3}&0&9\\0&3&1&{ - 2}\\0&0&2&0\\0&0&0&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3&0&0&0\\0&3&0&0\\0&0&3&0\\0&0&0&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}2&{ - 3}&0&9\\0&0&1&{ - 2}\\0&0&{ - 1}&0\\0&0&0&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

Therefore, the system of equations is shown below:

\(\begin{array}{c} - 2{x_1} - 3{x_2} + 9{x_4} = 0\\{x_3} - 2{x_4} = 0\\ - {x_3} = 0\\ - {x_4} = 0\end{array}\)

The general solution is,

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

Therefore, the eigenvector is \(\left\{ {{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\\0\\0\end{array}} \right)} \right\}\).

Write the matrix form for finding the eigenvector.

\(\begin{array}{c}\left( {A - 2I} \right){\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}5&{ - 3}&0&9\\0&3&1&{ - 2}\\0&0&2&0\\0&0&0&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}3&{ - 3}&0&9\\0&1&1&{ - 2}\\0&0&0&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

Therefore, the system of equations is shown below:

\(\begin{array}{c}3{x_1} - 3{x_2} + 9{x_4} = 0\\{x_2} + {x_3} - 2{x_4} = 0\end{array}\)

The general solution is,

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

Therefore, the eigenvectors are \(\left\{ {{{\rm{v}}_3},{{\rm{v}}_4}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\\1\end{array}} \right)} \right\}\).

The matrix\(P\)is formed by eigenvectors

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

The matrix\(D\)is formed by eigenvalues.

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

Thus, the matrix \(A\) is diagonalizable with \(P = \left( {\begin{array}{*{20}{c}}1&{\frac{3}{2}}&{ - 1}&{ - 1}\\0&1&{ - 1}&2\\0&0&1&0\\0&0&0&1\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}5&0&0&0\\0&3&0&0\\0&0&2&0\\0&0&0&2\end{array}} \right)\).