The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M } = \left( \begin{array} { r r r r } 1 & 1 & 5 & 7
3 & 9 & 17 & 25
1 & 7 & 7 & 11
3 & 6 & 16 & 23 \end{array} \right)\).
- In either order,
(a) show that the dimension of \(R\), the range space of T , is equal to 2 ,
(b) obtain a basis for \(R\). - Show that the vector \(\left( \begin{array} { r } 1
- 15
- 17
- 6 \end{array} \right)\) belongs to \(R\). - It is given that \(\left\{ \mathbf { e } _ { 1 } , \mathbf { e } _ { 2 } \right\}\) is a basis for the null space of T , where \(\mathbf { e } _ { 1 } = \left( \begin{array} { r } 14
1
- 3
0 \end{array} \right)\) and \(\mathbf { e } _ { 2 } = \left( \begin{array} { r } 19
2
0
- 3 \end{array} \right)\). Show that, for all \(\lambda\) and \(\mu\),
$$\mathbf { x } = \left( \begin{array} { r }
4
- 3
0
0
\end{array} \right) + \lambda \mathbf { e } _ { 1 } + \mu \mathbf { e } _ { 2 }$$
is a solution of
$$\mathbf { M x } = \left( \begin{array} { r }
1
- 15
- 17
- 6
\end{array} \right)$$ - Hence find a solution of \(( * )\) of the form \(\left( \begin{array} { c } \alpha
0
\gamma
\delta \end{array} \right)\).