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.

33. \(\left[ {\begin{array}{*{20}{c}}{ - {\bf{6}}}&{\bf{4}}&{\bf{0}}&{\bf{9}}\\{ - {\bf{3}}}&{\bf{0}}&{\bf{1}}&{\bf{6}}\\{ - {\bf{1}}}&{ - {\bf{2}}}&{\bf{1}}&{\bf{0}}\\{ - {\bf{4}}}&{\bf{4}}&{\bf{0}}&{\bf{7}}\end{array}} \right]\)

Let, \(A = \left[ {\begin{array}{*{20}{c}}{ - 6}&4&0&9\\{ - 3}&0&1&6\\{ - 1}&{ - 2}&1&0\\{ - 4}&4&0&7\end{array}} \right]\).

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

\(\begin{array}{l} > > A = \left[ {\begin{array}{*{20}{c}}{ - 6}&4&0&9\end{array};\,\,\begin{array}{*{20}{c}}{ - 3}&0&1&{6;\,\,\begin{array}{*{20}{c}}{ - 1}&{ - 2}&1&{0;\,\,\begin{array}{*{20}{c}}{ - 4}&4&0&7\end{array}}\end{array}}\end{array}} \right];\\ > > {\rm{ev}} = eigs\left( A \right);\end{array}\)

So, the eigenvalues of A are:

\(ev = \left( {5,1, - 2, - 2} \right)\)

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 \(\lambda = 5\) is:\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{1.0000}\\{0.500}\\{ - 0.5000}\\{1.0000}\end{array}} \right]\)

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

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

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

\({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{1.0000}\\{ - 0.5000}\\{ - 3.5000}\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = 1\) is \(\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\{ - 7}\\2\end{array}} \right]\).

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

So, the eigenvector for \(\lambda = - 2\) is:

\({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{1.0000}\\{1.0000}\\{1.0000}\\0\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}{1.5000}\\{ - 0.7500}\\0\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = - 2\) is \(\left[ {\begin{array}{*{20}{c}}1\\1\\1\\0\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}6\\{ - 3}\\0\\4\end{array}} \right]\).

The invertible matrix P can be written as,

\(P = \left[ {\begin{array}{*{20}{c}}2&2&1&6\\1&{ - 1}&1&{ - 3}\\{ - 1}&{ - 7}&1&0\\2&2&0&4\end{array}} \right]\)

The diagonalized form can be written as,

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

So, the diagonal form of the given matrix is \(\left[ {\begin{array}{*{20}{c}}5&0&0&0\\0&1&0&0\\0&0&{ - 2}&0\\0&0&0&{ - 2}\end{array}} \right]\).