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

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

Short answer

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

Find the eigenvalues and eigenvectors

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.