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

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{6}}&{{\bf{ - 1}}}\end{array}} \right)\)

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 \(A = \left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right)\). We need to diagonalize the given matrix if possible.

As the given matrix is an upper triangular matrix. Therefore, the eigenvalues of\(A\)are its diagonal values.

The eigenvalues are\({\lambda _1} = 1\)and\({\lambda _2} = - 1\).

Thus, the diagonal matrix is \(D = \left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)\).

Eigenvectors for \({\lambda _1} = 1\),

\[\begin{array}{c}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}0&0\\6&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Assume \({x_2} = t\)then we get:

\[\begin{array}{l}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}t\\{3t}\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1\\3\end{array}} \right)t\end{array}\]

Eigenvectors for \({\lambda _1} = - 1\),

\[\begin{array}{c}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}2&0\\6&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Assume \({x_2} = t\)then we get:

\[\begin{array}{l}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\t\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)t\end{array}\]

\[\begin{array}{c}P = \left\{ {{v_1},{v_2}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\end{array}\]

\[\begin{array}{l}{P^{ - 1}} = \frac{1}{{1\left( 1 \right) - 0}}\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right)\end{array}\]

\[\begin{array}{c}AP = \left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right)\end{array}\]

\[\begin{array}{c}PD = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right)\end{array}\]

Thus, as matrix product \(AP\) and \(PD\) are same, that is, the matrix \(A\) is diagonalizable.