Question

Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

5. \(\left( {\begin{array}{*{20}{c}}2&2&1\\1&3&1\\1&2&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&1&2\\1&0&{ - 1}\\1&{ - 1}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&0&0\\0&1&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{2}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{2}}&{ - \frac{3}{4}}\\{\frac{1}{4}}&{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)\)

Short answer

The basis for eigenvalue \(\lambda = 1\)is \(\left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right)\)and \(\left( {\begin{array}{*{20}{c}}2\\{ - 1}\\0\end{array}} \right)\)and the basis for eigenvalue \(\lambda = 5\)is \(\left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

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 the inverse of the invertible matrix

Write the given diagonalization matrix of \(A\) as:

\[\begin{array}{c}A = \left( {\begin{array}{*{20}{c}}2&2&1\\1&3&1\\1&2&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&1&2\\1&0&{ - 1}\\1&{ - 1}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&0&0\\0&1&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{2}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{2}}&{ - \frac{3}{4}}\\{\frac{1}{4}}&{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)\end{array}\].

Compare diagonalization matrix of \(A\) with \(A = PD{P^{ - 1}}\).

\[\begin{array}{c}P = \left( {\begin{array}{*{20}{c}}1&1&2\\1&0&{ - 1}\\1&{ - 1}&0\end{array}} \right)\\D = \left( {\begin{array}{*{20}{c}}5&0&0\\0&1&0\\0&0&1\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{2}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{2}}&{ - \frac{3}{4}}\\{\frac{1}{4}}&{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)\end{array}\]

Find Eigenvalues and Eigenvectors

According to the diagonal entries of the matrix, \(D\) there are three eigenvalues.

\({\lambda _1} = 5\), \({\lambda _2} = 1\)and\({\lambda _3} = 1\)

The three eigenvectors are columns of the matrix \(P\).

\[{{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right), {{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right), {{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\0\end{array}} \right)\]

Thus, the basis for eigenvalue \(\lambda = 1\)is \(\left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right)\)and \(\left( {\begin{array}{*{20}{c}}2\\{ - 1}\\0\end{array}} \right)\)and the basis for eigenvalue \(\lambda = 5\)is \(\left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).