Question

Show that the transformation in Exercise 8 is merely a rotation about the origin. What is the angle of the rotation?

Short answer

The angle of rotation is \(\frac{\pi }{2}\) radians.

Step-by-step solution

Find the value of \(T\) using linear transformation

Using linear transformation,

\(\begin{aligned} T &= T\left( {{x_1}{e_1} + {x_2}{e_2}} \right)\\ &= {x_1}T\left( {{e_1}} \right) + {x_2}T\left( {{e_2}} \right)\\ &= \left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}&{T\left( {{e_2}} \right)}\end{array}} \right]x\end{aligned}\)

Find the transformation \(T\) for \({e_1}\) and \({e_2}\)

In exercise 8, transformation \(T\)for \({e_1}\) is

\({e_1} \to {e_1} \to {e_2}\).

In exercise 8, transformation \(T\)for \({e_2}\) is

\({e_2} \to - {e_2} \to - {e_1}\).

The transformation matrix is

\(A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\).

Find the rotation about the origin

The transformation matrix \(\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\) can be written as follows:

\(\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\cos \frac{\pi }{2}}&{ - \sin \frac{\pi }{2}}\\{\sin \frac{\pi }{2}}&{\cos \frac{\pi }{2}}\end{array}} \right]\).

So, the angle of rotation is \(\frac{\pi }{2}\) radians.