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

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

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 \(A = \left( {\begin{array}{*{20}{c}}4&0&0&0\\0&4&0&0\\0&0&2&0\\1&0&0&2\end{array}} \right)\). Since it is given the eigenvalues of the matrix are \(4\), \(4\), \(2\) and \(2\).

Write the matrix form for finding the eigenvector.

\(\begin{array}{c}A - 4I = \left( {\begin{array}{*{20}{c}}0&0&0&0\\0&0&0&0\\0&0&{ - 2}&0\\1&0&0&{ - 2}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0&0&{ - 2}&0\\0&0&{ - 2}&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_1} \leftrightarrow {R_4}\\{R_2} \leftrightarrow {R_3}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&{ - 2}&0\\0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right)\;\left\{ {{R_2} = - \frac{{{R_2}}}{2}} \right\}\end{array}\)

The general solution is,

\(\left\{ {{{\rm{v}}_1},{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}0\\1\\0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\0\\0\\1\end{array}} \right)} \right\}\).

Therefore, the eigenvectors are \(\left\{ {{{\rm{v}}_1},{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}0\\1\\0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\0\\0\\1\end{array}} \right)} \right\}\).

Write the matrix form for finding the eigenvector.

\(\begin{array}{c}A - 2I = \left( {\begin{array}{*{20}{c}}2&0&0&0\\0&2&0&0\\0&0&0&0\\1&0&0&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&0&0&0\\1&0&0&0&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_1} \to \frac{{{R_1}}}{2}\\{R_2} \to \frac{{{R_2}}}{2}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right)\;\left\{ {{R_4} = {R_4} - {R_1}} \right\}\end{array}\)

The general solution is,

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

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

The matrix\(P\)is formed by eigenvectors

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

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

\(D = \left( {\begin{array}{*{20}{c}}4&0&0&0\\0&4&0&0\\0&0&2&0\\0&0&0&2\end{array}} \right)\)

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