Question 4.1.9: Let T: R³→ R³ be a linear operator and B a basis for R³ give...
Let T: R³→ R³ be a linear operator and B a basis for R³ given by
B= \left\{\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \right\}
If
T\left(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\right) = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \quad T\left(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\right) = \begin{bmatrix} -1 \\ -2 \\ -3 \end{bmatrix} \quad T\left(\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\right) = \begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}
find
T\left(\begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix}\right)
Since B is a basis for R³, there are (unique) scalars c_{1}, c_{2}, and c_{3} such that
c_{1} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} +c_{2}\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + c_{3} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix}
Solving this equation, we obtain c_{1} = −1, c_{2} = 1, and c_{3} = 2. Hence
T\left(\begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix}\right) = T \left(-1\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} +\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \right)
By the linearity of T, we have
T\left(\begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix}\right) = (-1) T\left(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\right) + T\left(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\right)+2 T\left(\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\right)
= – \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} +\begin{bmatrix} -1 \\ -2 \\ -3 \end{bmatrix}+2 \begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}
= \begin{bmatrix} 2 \\ 1 \\ 4 \end{bmatrix}