Question

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

15. \(\left( {\begin{array}{*{20}{c}}{\bf{7}}&{\bf{4}}&{{\bf{16}}}\\{\bf{2}}&{\bf{5}}&{\bf{8}}\\{{\bf{ - 2}}}&{{\bf{ - 2}}}&{{\bf{ - 5}}}\end{array}} \right)\)

Short answer

The matrix \(A\) is diagonalizable with \(P = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 4}&{ - 2}\\1&0&{ - 1}\\0&1&1\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}3&0&0\\0&3&0\\0&0&1\end{array}} \right)\).

Step-by-step solution

Write the Diagonalization Theorem

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 \(A\).

Find eigenvalues of the matrix

Consider the given matrix \(A = \left( {\begin{array}{*{20}{c}}7&4&{16}\\2&5&8\\{ - 2}&{ - 2}&{ - 5}\end{array}} \right)\). Since from the given matrix eigenvalues of the matrix are \(3\) and \(1\). Since the eigenvalues are distinct, the matrix is diagonalizable.

Find the eigenvalues and eigenvectors

As the sum of all the eigenvalues of\(A\)is equal to the sum of the diagonal entries of\(A\), we can find the third eigenvalue:

\(\begin{array}{c}3 + 1 + x = 7 + 5 - 5\\x = 3\end{array}\)

So, the eigenvalues are\(3,3,1\).

Find eigenvectors.

Write the matrix form for finding the eigenvector for\(\lambda = 3\).

\(\begin{array}{c}A - 3I = 0\\\left( {\begin{array}{*{20}{c}}7&4&{16}\\2&5&8\\{ - 2}&{ - 2}&{ - 5}\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}4&4&{16}\\2&2&8\\{ - 2}&{ - 2}&{ - 8}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\end{array}\)

Write the Row-reduced Augmented matrix.

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}4&4&{16}&0\\2&2&8&0\\{ - 2}&{ - 2}&{ - 8}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&1&4&0\\2&2&8&0\\{ - 2}&{ - 2}&{ - 8}&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_1} = \frac{{{R_1}}}{4}\\{R_2} = {R_2} - 2{R_1}\\{R_3} = {R_3} + 2{R_1}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&1&4&0\\0&0&0&0\\0&0&0&0\end{array}} \right)\end{array}\)

Therefore, the parametric form of the solution is shown below:

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - {x_2} - 4{x_3}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - {x_2}}\\{{x_2}}\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 4{x_3}}\\0\\{{x_3}}\end{array}} \right)\\ = {x_2}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 4}\\0\\1\end{array}} \right)\end{array}\)

Therefore, the eigenvector for \(\lambda = 3\)are\(\left\{ {{{\rm{v}}_1},{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 1}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 4}\\0\\1\end{array}} \right)} \right\}\).

Find eigenvectors.

Write the matrix form for finding the eigenvector for\(\lambda = 1\).

\(\begin{array}{c}A - I = 0\\\left( {\begin{array}{*{20}{c}}7&4&{16}\\2&5&8\\{ - 2}&{ - 2}&{ - 5}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}6&4&{16}\\2&4&8\\{ - 2}&{ - 2}&{ - 6}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\end{array}\)

Write the Row-reduced Augmented matrix.

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}6&4&{16}&0\\2&4&8&0\\{ - 2}&{ - 2}&{ - 6}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&{\frac{2}{3}}&{\frac{8}{3}}&0\\2&{\frac{8}{3}}&{\frac{8}{3}}&0\\{ - 2}&{ - \frac{2}{3}}&{ - \frac{2}{3}}&0\end{array}} \right)\;\left\{ {{R_1} = \frac{{{R_1}}}{6}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&{\frac{2}{3}}&{\frac{8}{3}}&0\\0&{\frac{8}{3}}&{\frac{8}{3}}&0\\0&{ - \frac{2}{3}}&{ - \frac{2}{3}}&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_2} = {R_2} - 2{R_1}\\{R_3} = {R_3} + 2{R_1}\\{R_1} = {R_1} + {R_3}\\{R_2} = \frac{3}{8}{R_2}\\{R_3} = {R_3} + {R_2}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&2&0\\0&1&1&0\\0&0&0&0\end{array}} \right)\end{array}\)

Therefore, the parametric form of the solution is shown below:

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

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

Find the matrix \(P\) and \(D\)

The matrix\(P\)is formed by eigenvectors

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

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

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

Find the diagonalizes form of a matrix \(A\)

As the diagonal form of the matrix\(A\)is\(A = PD{P^{ - 1}}\).

Where\(P = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 4}&{ - 2}\\1&0&{ - 1}\\0&1&1\end{array}} \right)\)and\(D = \left( {\begin{array}{*{20}{c}}3&0&0\\0&3&0\\0&0&1\end{array}} \right)\).

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