A linear transformation \(T:{\mathbb{R}^2} \to {\mathbb{R}^2}\), first reflects points through the \({x_1}\)-axis and then reflects points through the \({x_2}\)-axis. Show that \(T\) can also be described as a linear transformation that rotates points about the origin. What is the angle of that rotation?

Using linear transformation,

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

For \({e_1}\), when it is reflected through the horizontal \({x_1}\)-axis, then

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

When it is reflected through \({x_2}\)-axis,

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

For \({e_2}\), when it is reflected through the horizontal \({x_1}\)-axis, then

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

When it is reflected through the line \({x_2} = - {x_1}\),

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

By the equation \(T = \left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}&{T\left( {{e_2}} \right)}\end{array}} \right]x\),

\(\begin{aligned} T &= \left[ {\begin{array}{*{20}{c}}{ - {e_1}}&{ - {e_2}}\end{array}} \right]x\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]x\end{aligned}\)

By the equation \(T = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]x\), the matrix \(A = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]\).

So, thelinear transformation matrix is \(\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]\).

The linear transformation is completely determined by what it does to the columns of the identity matrix. The rotation transformation has the same effect as \(T\) on every vector \({\mathbb{R}^2}\).

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

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

So, the angle of rotation is \(\pi \) radians.