Question: 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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

10. \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{1}}\end{array}} \right)\)

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 \(A = \left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right)\). We need to diagonalize the given matrix if possible.

Write the characteristic equation:

\[\begin{array}{c}\left| {A - \lambda I} \right| = 0\\\left| {\begin{array}{*{20}{c}}{2 - \lambda }&3\\4&{1 - \lambda }\end{array}} \right| = 0\\\left( {\lambda + 2} \right)\left( {\lambda - 5} \right) = 0\\\lambda = - 2,5\end{array}\]

Write the matrix form for finding the eigenvector for \(\lambda = - 2\).

\[\begin{array}{c}\left( {A - \lambda I} \right){\bf{x}} = 0\\\left( {A - \left( { - 2} \right)I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}4&3\\4&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Therefore, the augmented matrix is shown below:

\[\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}4&3&0\\4&3&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4&3&0\\4&3&0\end{array}} \right)\;\left\{ {{R_2} = {R_2} - {R_1}} \right\}\end{array}\]

Since there are \(2\) variables and \(1\) equitation, consider \({x_2}\) as free.

\[\begin{array}{c}4{x_1} + 3{x_2} = 0\\4{x_1} = - 3{x_2}\\{x_1} = - \frac{3}{4}{x_2}\end{array}\]

Write the parametric form of solution.

\[\begin{array}{c}{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\\ = {x_2}\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}\\1\end{array}} \right)\end{array}\]

Therefore, the eigenvector for \(\lambda = - 2\) is \({\rm{v}} = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}\\1\end{array}} \right)\).

Write the matrix form for finding the eigenvector for \(\lambda = 5\).

\[\begin{array}{c}\left( {A - \lambda I} \right){\bf{x}} = 0\\\left( {A - \left( 5 \right)I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{ - 3}&3\\4&{ - 4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Therefore, the augmented matrix is shown below:

\[\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}{ - 3}&3&0\\4&{ - 4}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&{ - 1}&0\\1&{ - 1}&0\end{array}} \right)\;\left\{ {{R_1} = - \frac{{{R_1}}}{3},{R_2} = - \frac{{{R_2}}}{3}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&{ - 1}&0\\0&0&0\end{array}} \right)\;\left\{ {{R_1} = {R_2} - {R_1}} \right\}\end{array}\]

Since there are \(2\) variables and \(1\) equitation, consider \({x_2}\) as free.

\[\begin{array}{c}{x_1} - {x_2} = 0\\{x_1} = {x_2}\end{array}\]

Write the parametric form of solution.

\[\begin{array}{c}{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\\ = {x_2}\left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)\end{array}\]

Therefore, the eigenvector for \(\lambda = 5\) is \(v = \left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)\).

The matrix \(P\) is formed by eigenvectors

\[P = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)\]

The inverse of the matrix \(P\) is shown below:

\[\begin{array}{c}{P^{ - 1}} = {\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)^{ - 1}}\\ = \frac{1}{{1\left( { - \frac{3}{4}} \right) - 1}}\left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&{ - \frac{3}{4}}\end{array}} \right)\\ = - \frac{4}{7}\left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&{ - \frac{3}{4}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - \frac{4}{7}}&{\frac{4}{7}}\\{\frac{4}{7}}&{\frac{3}{7}}\end{array}} \right)\end{array}\]

As the matrix that diagonalizes \(A\)is shown below:

\[\begin{array}{c}D = {P^{ - 1}}AP\\ = \left( {\begin{array}{*{20}{c}}{ - \frac{4}{7}}&{\frac{4}{7}}\\{\frac{4}{7}}&{\frac{3}{7}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{8}{7}}&{ - \frac{8}{7}}\\{\frac{{20}}{7}}&{\frac{{15}}{7}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&5\end{array}} \right)\end{array}\]

Thus, the matrix \(A\) is diagonalizable with \(P = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)\), \({P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - \frac{4}{7}}&{\frac{4}{7}}\\{\frac{4}{7}}&{\frac{3}{7}}\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&5\end{array}} \right)\).