Question

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

Short answer

\(P = \frac{1}{2}\left( {\begin{aligned}{{}}{ - 1}&1&{ - 1}&1\\1&1&{ - 1}&{ - 1}\\{ - 1}&1&1&{ - 1}\\1&1&1&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{{}}{19}&0&0&0\\0&{11}&0&0\\0&0&5&0\\0&0&0&{ - 11}\end{aligned}} \right)\)

Step-by-step solution

Step 1:Find the eigenvalues of the matrix

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\(E = \left( {\begin{aligned}{{}}{19}\\{11}\\5\\{ - 11}\end{aligned}} \right)\).

Find the eigenvectors of the matrix

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 \({v_4} = \left( {\begin{aligned}{{}}1\\{ - 1}\\{ - 1}\\1\end{aligned}} \right)\)

Find the orthogonal projection

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,

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

Find the matrix P and D

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\(D = \left( {\begin{aligned}{{}}{19}&0&0&0\\0&{11}&0&0\\0&0&5&0\\0&0&0&{ - 11}\end{aligned}} \right)\).