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

11. \(\left( {\begin{array}{*{20}{c}}{ - 1}&4&{ - 2}\\{ - 3}&4&0\\{ - 3}&1&3\end{array}} \right)\)

Short answer

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

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 \(A = \left( {\begin{array}{*{20}{c}}{ - 1}&4&{ - 2}\\{ - 3}&4&0\\{ - 3}&1&3\end{array}} \right)\). Since from the given matrix eigenvalues of the matrix are \(1\), \(2\) and \(3\). Since the eigenvalues are distinct, the matrix is diagonalizable.

Find the eigenvectors

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

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

Write the parametric form of solution.

\(\begin{array}{c} - 2{x_1} + 4{x_2} - 2{x_3} = 0\\ - 3{x_1} + 3{x_2} = 0\\ - 3{x_1} + {x_2} + 2{x_3} = 0\end{array}\)

On solving we get:

\(\left\{ \begin{array}{l}{x_1} = {x_2}\\{x_1} = {x_3}\\{x_3} = r\end{array} \right\}\)

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

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

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

Write the parametric form of solution.

\(\begin{array}{c} - 2{y_1} + 4{y_2} - 2{y_3} = 0\\ - 3{y_1} + 2{y_2} = 0\\ - 3{y_1} + {y_2} + {y_3} = 0\end{array}\)

On solving we get:

\(\left\{ \begin{array}{l}{y_1} = \frac{2}{3}{y_2}\\{y_2} = {y_3}\\{y_3} = r\end{array} \right\}\)

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

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

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

Write the parametric form of solution.

\(\begin{array}{c} - 4{z_1} + 4{z_2} - 2{z_3} = 0\\ - 3{z_1} + {z_2} = 0\end{array}\)

On solving we get:

\(\left\{ \begin{array}{l}{z_1} = r\\{z_2} = 3{z_1}\\{z_3} = 4{z_1}\end{array} \right\}\)

Therefore, the eigenvector for \(\lambda = 3\) is \({{\rm{v}}_3} = \left\{ {\left( {\begin{array}{*{20}{c}}\begin{array}{l}1\\3\end{array}\\4\end{array}} \right)} \right\}\).

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

The matrix\(P\)is formed by eigenvectors

\(P = \left( {\begin{array}{*{20}{c}}1&{\frac{2}{3}}&1\\1&1&3\\1&1&4\end{array}} \right)\)

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

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

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

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

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

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