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.

34. \(\left[ {\begin{array}{*{20}{c}}{\bf{0}}&{{\bf{13}}}&{\bf{8}}&{\bf{4}}\\{\bf{4}}&{\bf{9}}&{\bf{8}}&{\bf{4}}\\{\bf{8}}&{\bf{6}}&{{\bf{12}}}&{\bf{8}}\\{\bf{0}}&{\bf{5}}&{\bf{0}}&{ - {\bf{4}}}\end{array}} \right]\)

Let,

\[A = \left[ {\begin{array}{*{20}{c}}0&{13}&8&4\\4&9&8&4\\8&6&{12}&8\\0&5&0&{ - 4}\end{array}} \right]\]

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

\(\begin{array}{l} > > A = \left[ {\begin{array}{*{20}{c}}0&{13}&8&4\end{array};\,\,\begin{array}{*{20}{c}}4&9&8&{4;\,\,\begin{array}{*{20}{c}}8&6&{12}&{8;\,\,\begin{array}{*{20}{c}}0&5&0&{ - 4}\end{array}}\end{array}}\end{array}} \right];\\ > > {\rm{ev}} = eigs\left( A \right);\end{array}\)

So, the eigenvalues of A are:

\(ev = \left( { - 4,24,1, - 4} \right)\)

Find the eigenvalues of A using the following MATLAB code:

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 1 \right) * {\rm{eye}}\left( 4 \right)} \right);\)

So, the eigenvector for \(\lambda = - 4\) is:

\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\1\\0\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}{ - 1}\\0\\0\\1\end{array}} \right]\)

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

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 2 \right) * {\rm{eye}}\left( 4 \right)} \right);\)

So, the eigenvector for \(\lambda = 24\) is:

\({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{5.6000}\\{5.6000}\\{7.2000}\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = 24\) is \(\left[ {\begin{array}{*{20}{c}}{28}\\{28}\\{36}\\5\end{array}} \right]\).

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 3 \right) * {\rm{eye}}\left( 4 \right)} \right);\)

So, the eigenvector for \(\lambda = 1\) is:

\({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{1.0000}\\{1.0000}\\{ - 2.0000}\\{1.0000}\end{array}} \right]\)

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

The invertible matrix P can be written as,

\(P = \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 1}&{28}&1\\0&0&{28}&1\\1&0&{36}&{ - 2}\\0&1&5&1\end{array}} \right]\)

The diagonalized form can be written as,

\(D = \left[ {\begin{array}{*{20}{c}}{ - 4}&0&0&0\\0&{ - 4}&0&0\\0&0&{24}&0\\0&0&0&1\end{array}} \right]\)

So, the diagonal form of the given matrix is \(\left[ {\begin{array}{*{20}{c}}{ - 4}&0&0&0\\0&{ - 4}&0&0\\0&0&{24}&0\\0&0&0&1\end{array}} \right]\).