Question 4.4.1: Define the linear operator T: R³ → R³ by T([ x y z]) = [ x –...
Define the linear operator T: R³ → R³ by
T \left(\begin{bmatrix} x\\ y\\ z \end{bmatrix} \right) =\begin{bmatrix} x \\ -y \\ z \end{bmatrix}
a. Find the matrix of T relative to the standard basis for R³.
b. Use the result of part (a) to find
T \left(\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix} \right)
a. Let B = \left\{e_{1}, e_{2}, e_{3}\right\} . be the standard basis for R³. Since
\left[T \left(e_{1} \right) \right] _{B} = \begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix} \left[T \left(e_{2} \right) \right] _{B} = \begin{bmatrix} 0 \\ -1\\ 0 \end{bmatrix} \left[T \left(e_{3} \right) \right] _{B} = \begin{bmatrix} 0 \\ 0\\ 1 \end{bmatrix}.
then
\left[T\right] _{B} =\begin{bmatrix} 1&0&0 \\ 0&-1&0\\ 0&0&1 \end{bmatrix} .
b. Since B is the standard basis for R³, the coordinates of any vector are given by its components. In this case, with
v=\begin{bmatrix} 1 \\ 1\\ 2 \end{bmatrix} then \left[V\right] _{B} = \begin{bmatrix} 1 \\ 1\\ 2 \end{bmatrix}.
Thus, by Theorem 12,
T (v) = \left[T(v)\right] _{B} \begin{bmatrix} 1&0&0 \\ 0&-1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1 \\ 1\\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ -1\\ 2 \end{bmatrix}.
Notice that the action of T is a reflection through the xz plane, as shown in x Fig. 1.
