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

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

Short answer

The general solution is \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\). Since we cannot generate an eigenvector basis for \({\mathbb{R}^2}\), \(A\) is not diagonalizable.

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 \(A = \left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)\). We need to diagonalize the given matrix if possible.

Write the characteristic equation:

\[\begin{array}{c}\det \left( {A - \lambda I} \right) = \det \left( {\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right) - \lambda \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\\ = \det \left( {\left( {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 1}\\1&{5 - \lambda }\end{array}} \right)} \right)\\ = \left( {3 - \lambda } \right)\left( {5 - \lambda } \right) - \left( { - 1} \right)\left( 1 \right)\\ = {\lambda ^2} - 8\lambda + 16\\ = {\left( {\lambda - 4} \right)^2}\end{array}\]

Therefore, \(\det \left( {A - \lambda I} \right) = {\left( {\lambda - 4} \right)^2}\).

Find the eigenvalues

\[\begin{array}{c}\det \left( {A - \lambda I} \right) = 0\\{\left( {\lambda - 4} \right)^2} = 0\\\lambda = 4,4\end{array}\]

So, the eigenvalue of the matrix is \(4\) with a multiplicity of \(2\).

Find the eigenvectors

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

Therefore, the eigenvectors are shown below:

\[ - {x_1} - {x_2} = 0\]

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

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

Since we cannot generate an eigenvector basis for \({\mathbb{R}^2}\), \(A\) is not diagonalizable.

Thus, the matrix is not-Diagonalizable.