Question: Diagonalize the matrices in Exercises 33-36. Use your matrix program’s eigenvalue command to find the eigenvalues, and then compute bases for the eigenspaces as in section 5.1.

36. \(\left[ {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{4}}&{\bf{2}}&{\bf{3}}&{ - {\bf{2}}}\\{\bf{0}}&{\bf{1}}&{ - {\bf{2}}}&{ - {\bf{2}}}&{\bf{2}}\\{\bf{6}}&{{\bf{12}}}&{{\bf{11}}}&{\bf{2}}&{ - {\bf{4}}}\\{\bf{9}}&{{\bf{20}}}&{{\bf{10}}}&{{\bf{10}}}&{ - {\bf{6}}}\\{{\bf{15}}}&{{\bf{28}}}&{{\bf{14}}}&{\bf{5}}&{ - {\bf{3}}}\end{array}} \right]\)

Let,

\[A = \left[ {\begin{array}{*{20}{c}}4&4&2&3&{ - 2}\\0&1&{ - 2}&{ - 2}&2\\6&{12}&{11}&2&{ - 4}\\9&{20}&{10}&{10}&{ - 6}\\{15}&{28}&{14}&5&{ - 3}\end{array}} \right]\]

Use the following code in MATLAB to find the eigenvalues of the matrix.

\[\begin{array}{l} > > A = \left[ \begin{array}{c}\begin{array}{*{20}{c}}4&4&2&3&{ - 2;\,}\end{array}\,\begin{array}{*{20}{c}}0&1&{ - 2}&{ - 2}&{ - 2;\,}\end{array}\,\begin{array}{*{20}{c}}6&{12}&{11}&2&{ - 4;\,\,}\end{array}\,\\\begin{array}{*{20}{c}}9&{20}&{10}&{10}&{ - 6;\,}\end{array}\begin{array}{*{20}{c}}{15}&{28}&{14}&5&{ - 3}\end{array}\end{array} \right];\\ > > {\rm{ev}} = eigs\left( A \right);\end{array}\]

So, the eigenvalues of A are:

\(ev = \left( {3,5,7,5,3} \right)\)

Find the eigenvalues of A using the following MATLAB code:

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 1 \right) * {\rm{eye}}\left( 5 \right)} \right);\)

So, the eigenvector for \(\lambda = 3\) is:

\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{2.0000}\\{ - 1.5000}\\{0.5000}\\{1.0000}\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 1.0000}\\{0.5000}\\{0.5000}\\0\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = 3\) is \(\left[ {\begin{array}{*{20}{c}}4\\{ - 3}\\1\\2\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\1\\0\\2\end{array}} \right]\).

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 2 \right) * {\rm{eye}}\left( 5 \right)} \right);\)

So, the eigenvector for \(\lambda = 5\) is:

\({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 0.5000}\\{1.0000}\\0\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 1.0000}\\{1.0000}\\0\\{ - 1.0000}\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = 5\) is \(\left[ {\begin{array}{*{20}{c}}0\\{ - 1}\\2\\0\\0\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}{ - 1}\\1\\0\\{ - 1}\\1\end{array}} \right]\).

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 3 \right) * {\rm{eye}}\left( 5 \right)} \right);\)

So, the eigenvector for \(\lambda = 7\) is:

\({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{0.3300}\\{0.0000}\\{0.0000}\\{1.0000}\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = 3\) is \(\left[ {\begin{array}{*{20}{c}}1\\0\\0\\3\\4\end{array}} \right]\).

The invertible matrix P can be written as,

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

The diagonalized form can be written as,

\(D = \left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\0&3&0&0&0\\0&0&5&0&0\\0&0&0&5&0\\0&0&0&0&7\end{array}} \right]\)

So, the diagonal form of the given matrix is \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\0&3&0&0&0\\0&0&5&0&0\\0&0&0&5&0\\0&0&0&0&7\end{array}} \right]\).