(M) Let \(T\) be the transformation whose standard matrix is given below. Find a basis for \({\mathbb{R}^4}\) with the property that \({\left( T \right)_B}\) is diagonal.

\(A = \left( {\begin{aligned}{}{15}&{}&{ - 66}&{}&{ - 44}&{}&{ - 33}\\0&{}&{13}&{}&{21}&{}&{ - 15}\\1&{}&{ - 15}&{}&{ - 21}&{}&{12}\\2&{}&{ - 18}&{}&{ - 22}&{}&8\end{aligned}} \right)\)

Use matrix method to find the Eigen values of the given matrix \(A\), by following steps given below:

1. Enter the matrix \(A\) into the program.
2. Apply the command, \({\rm{ev}}\,{\rm{ = }}\,{\rm{eign}}\left( {\rm{A}} \right)\).
3. Press Enter.

On running the command, we obtain the following result:

\(\begin{aligned}{}{\rm{ev}}\,{\rm{ = }}\,{\rm{eign}}\left( {\rm{A}} \right)\\ = \left( {2,\,\,4,\,\,4,\,\,5} \right)\end{aligned}\)

Use matrix method to find the Null basis for Eigen value, \(\lambda = 2\) of the given matrix \(A\), by following steps given below:

1. Apply the command, \({\rm{nullbasis}}\left( {{\rm{A - ev}}\left( {\rm{1}} \right)} \right)*\,{\rm{eye}}\left( 4 \right)\).
2. Press Enter.

On running the command, we obtain the following result:

\({\rm{nullbasis}}\left( {{\rm{A - ev}}\left( 2 \right)} \right)*\,{\rm{eye}}\left( 4 \right) = \left( \begin{aligned}{l} - 10.0000\\\,\, - 2.3333\\\,\,\,\,\,\,1.0000\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\end{aligned} \right)\left( \begin{aligned}{l}13.0000\\\,1.6667\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\\,\,1.0000\end{aligned} \right)\). Multiply it by 3 to obtain the basis for Eigen space of \(\lambda = 4\), as follows:

\(\left\{ {{{\bf{b}}_2},{{\bf{b}}_3}} \right\} = \left\{ {\left( \begin{aligned}{l} - 30\\\, - 7\\\,\,\,\,3\\\,\,\,\,0\end{aligned} \right)\left( \begin{aligned}{l}39\\5\\\,0\\\,3\end{aligned} \right)} \right\}\).

Following the similar steps to find Null basis for Eigen value, \(\lambda = 4\) of the given matrix \(A\), as follows:

\({\rm{nullbasis}}\left( {{\rm{A - ev}}\left( {\rm{1}} \right)} \right)*\,{\rm{eye}}\left( 4 \right) = \left( \begin{aligned}{l}\,\,\,0.0000\\ - 1.5000\\\,\,\,\,1.5000\\\,\,\,\,1.0000\end{aligned} \right)\). Multiply it by 2 to obtain the basis for Eigen space of \(\lambda = 2\), as follows:

\({{\bf{b}}_1} = \left( \begin{aligned}{l}\,\,0\\ - 3\\\,\,\,3\\\,\,\,2\end{aligned} \right)\).

Following the similar steps to find Null basis for Eigen value, \(\lambda = 5\) of the given matrix \(A\), as follows:

\({\rm{nullbasis}}\left( {{\rm{A - ev}}\left( 4 \right)} \right)*\,{\rm{eye}}\left( 4 \right) = \left( \begin{aligned}{l}\,\,\,2.7000\\ - 0.7500\\\,\,\,\,1.0000\\\,\,\,\,1.0000\end{aligned} \right)\). Multiply it by 4 to obtain the basis for Eigen space of \(\lambda = 5\), as follows:

\({{\bf{b}}_4} = \left( \begin{aligned}{l}\,\,11\\ - 3\\\,\,\,4\\\,\,\,4\end{aligned} \right)\).

The basis \(B\),for \({\mathbb{R}^4}\) such that \({\left( T \right)_B}\) is diagonal is \(B = \left\{ {{{\bf{b}}_1},\,{{\bf{b}}_2},\,{{\bf{b}}_3},\,{{\bf{b}}_4}} \right\}\), which is written as follows:

\(B = \left( {\begin{aligned}{}0&{}&{ - 30}&{}&{39}&{}&{11}\\{ - 3}&{}&{ - 7}&{}&5&{}&{ - 3}\\3&{}&3&{}&0&{}&4\\2&{}&0&{}&3&{}&4\end{aligned}} \right)\)