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

17. \(\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{4}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{5}}\end{array}} \right)\)

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 \(A = \left( {\begin{array}{*{20}{c}}4&0&0\\1&4&0\\0&0&5\end{array}} \right)\). Since from the given matrix eigenvalues of the matrix are \(2\) and \(1\). Since the eigenvalues are distinct, the matrix is diagonalizable.

Write the characteristic equation:

\(\begin{array}{c}2 + 1 + x = 0 + 0 + 5\\x = 2\end{array}\)

So, the eigenvalues are\(2,2,1\).

Find eigenvectors.

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

\(\begin{array}{c}A - \lambda I = \left( {\begin{array}{*{20}{c}}4&0&0\\1&4&0\\0&0&5\end{array}} \right) - \lambda \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 - \lambda }&0&0\\1&{4 - \lambda }&0\\0&0&{5 - \lambda }\end{array}} \right)\end{array}\)

Now solve the characteristic polynomial

\(\begin{array}{c}\det \left( {A - \lambda I} \right) = 0\\\left( {4 - \lambda } \right)\left( {4 - \lambda } \right)\left( {5 - \lambda } \right) = 0\\\lambda = 4,4,5\end{array}\)

So, the eigenvalues are \(4,4,5\).

Write the matrix equation for finding the eigenvector\(\lambda = 4\).

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

Therefore, the general solution is shown below:

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

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

Since the basis for Eigenspace corresponding to\(\lambda = 4\)is\(\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\), therefore, the dimension of Eigenspace is one.

Hence the matrix \(A\) is not diagonalizable.