Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

37. \(\left( {\begin{aligned}{{}}{\bf{6}}&{\bf{2}}&{\bf{9}}&{ - {\bf{6}}}\\{\bf{2}}&{\bf{6}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{9}}&{ - {\bf{6}}}&{\bf{6}}&{\bf{2}}\\{\bf{6}}&{\bf{9}}&{\bf{2}}&{\bf{6}}\end{aligned}} \right)\)

Use the following MATLAB code to find the eigenvalues of the given matrix:

\(\begin{aligned}{} > > A = \left( {\begin{aligned}{{}}6&2&9&{ - 6}\end{aligned};\,\begin{aligned}{{}}2&6&{ - 6}&9\end{aligned};\,\begin{aligned}{{}}9&{ - 6}&6&2\end{aligned};\,\begin{aligned}{{}}{ - 6}&9&2&6\end{aligned}} \right);\\ > > \left( {\begin{aligned}{{}}{\rm{E}}&{\rm{V}}\end{aligned}} \right) = {\rm{eigs}}\left( A \right);\end{aligned}\)

So, the eigenvalues are\(E = \left( {\begin{aligned}{{}}{19}\\{11}\\5\\{ - 11}\end{aligned}} \right)\).

Use the following MATLAB code to find eigenvectors.

\( > > {v_i} = {\rm{nullbasis}}\left( {A - E\left( i \right)*{\rm{eye}}\left( 4 \right)} \right)\)

Following are the eigenvectors of A.

\({v_1} = \left( {\begin{aligned}{{}}{ - 1}\\1\\{ - 1}\\1\end{aligned}} \right)\), \({v_2} = \left( {\begin{aligned}{{}}1\\1\\1\\1\end{aligned}} \right)\), \({v_3} = \left( {\begin{aligned}{{}}{ - 1}\\{ - 1}\\1\\1\end{aligned}} \right)\), and \({v_4} = \left( {\begin{aligned}{{}}1\\{ - 1}\\{ - 1}\\1\end{aligned}} \right)\)

The orthogonal projections can be calculated as follows:

\(\begin{aligned}{}{{\bf{u}}_1} &= \frac{1}{{\left\| {{v_1}} \right\|}}{v_1}\\ &= \frac{1}{2}\left( {\begin{aligned}{{}}{ - 1}\\1\\{ - 1}\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{ - \frac{1}{2}}\\{\frac{1}{2}}\\{ - \frac{1}{2}}\\{\frac{1}{2}}\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{u}}_2} &= \frac{1}{{\left\| {{v_2}} \right\|}}{v_2}\\ &= \frac{1}{2}\left( {\begin{aligned}{{}}1\\1\\1\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{1}{2}}\\{\frac{1}{2}}\\{\frac{1}{2}}\\{\frac{1}{2}}\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{u}}_3} &= \frac{1}{{\left\| {{v_3}} \right\|}}{v_3}\\ &= \frac{1}{2}\left( {\begin{aligned}{{}}{ - 1}\\{ - 1}\\1\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{ - \frac{1}{2}}\\{ - \frac{1}{2}}\\{\frac{1}{2}}\\{\frac{1}{2}}\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{u}}_4} &= \frac{1}{{\left\| {{v_4}} \right\|}}{v_4}\\ &= \frac{1}{2}\left( {\begin{aligned}{{}}1\\{ - 1}\\{ - 1}\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{1}{2}}\\{ - \frac{1}{2}}\\{ - \frac{1}{2}}\\{\frac{1}{2}}\end{aligned}} \right)\end{aligned}\)

The matrix P can be written using orthogonal projections:

\(\begin{aligned}{}P &= \left( {\begin{aligned}{{}}{ - \frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\end{aligned}} \right)\\ &= \frac{1}{2}\left( {\begin{aligned}{{}}{ - 1}&1&{ - 1}&1\\1&1&{ - 1}&{ - 1}\\{ - 1}&1&1&{ - 1}\\1&1&1&1\end{aligned}} \right)\end{aligned}\)

The diagonalized matrix can be written as\(D = \left( {\begin{aligned}{{}}{19}&0&0&0\\0&{11}&0&0\\0&0&5&0\\0&0&0&{ - 11}\end{aligned}} \right)\).