Question: Show that columns of matrix A are orthogonal by making an appropriate matrix calculation. State the calculation you use.

\(A = \left( {\begin{array}{*{20}{c}}{ - {\bf{6}}}&{ - {\bf{3}}}&{\bf{6}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{2}}&{\bf{1}}&{ - {\bf{6}}}\\{\bf{3}}&{\bf{6}}&{\bf{3}}&{ - {\bf{2}}}\\{\bf{6}}&{ - {\bf{3}}}&{\bf{6}}&{ - {\bf{1}}}\\{\bf{2}}&{ - {\bf{1}}}&{\bf{2}}&{\bf{3}}\\{ - {\bf{3}}}&{\bf{6}}&{\bf{3}}&{\bf{2}}\\{ - {\bf{2}}}&{ - {\bf{1}}}&{\bf{2}}&{ - {\bf{3}}}\\{\bf{1}}&{\bf{2}}&{\bf{1}}&{\bf{6}}\end{array}} \right)\)

Use the following MATLAB command to obtain the transpose of matrix A.

\(\begin{array}{c} > > {\rm{A}} = \left( \begin{array}{l}\begin{array}{*{20}{c}}{ - 6}&{ - 3}&6&1\end{array};\,\begin{array}{*{20}{c}}{ - 1}&2&1&{ - 6}\end{array};\,\begin{array}{*{20}{c}}3&6&3&{ - 2;\,\begin{array}{*{20}{c}}6&{ - 3}&6&{ - 1}\end{array}}\end{array};\\\begin{array}{*{20}{c}}2&{ - 1}&2&{3;\,\begin{array}{*{20}{c}}{ - 3}&6&3&{2;\,\,\begin{array}{*{20}{c}}{ - 2}&{ - 1}&2&{ - 3;\,\,\begin{array}{*{20}{c}}1&2&1&6\end{array}}\end{array}}\end{array}}\end{array}\end{array} \right);\\ > > {\rm{A'}} = {\rm{A}};\end{array}\)

The transpose of matrix A is shown below:

\({A^T} = \left( {\begin{array}{*{20}{c}}{ - 6}&{ - 1}&3&6&2&{ - 3}&{ - 2}&1\\{ - 3}&2&6&{ - 3}&{ - 1}&6&{ - 1}&2\\6&1&3&6&2&3&2&1\\1&{ - 6}&{ - 2}&{ - 1}&3&2&{ - 3}&6\end{array}} \right)\)

The product \({A^T}*A\) can be calculated by using the following MATLAB code:

\({\rm{P}} = {\rm{A'}}*{\rm{A}}\)

The product \({A^T}*A\) is:

\(\begin{array}{c}{A^T}A = \left( {\begin{array}{*{20}{c}}{100}&0&0&0\\0&{100}&0&0\\0&0&{100}&0\\0&0&0&{100}\end{array}} \right)\\ = 100\left( {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}} \right)\\ = 100{I_4}\end{array}\)

As the matrix \({A^T}A\) has only diagonal entries, so the columns of matrix A are orthogonal.