In Exercise 1-10, assume that \(T\) is a linear transformation. Find the standard matrix of \(T\).

\(T:{\mathbb{R}^2} \to {\mathbb{R}^2}\), is a vertical shear transformation that maps \({e_1}\) into \({e_1} - 2{e_2}\) but leaves the vector \({e_2}\) unchanged.

Usinglinear transformation,

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

Here,

\(\begin{aligned} T\left( {{e_1}} \right) &= {e_1} - 2{e_2}\\ &= \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right]\end{aligned}\)

And

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

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

\(T = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right]x\).

By the equation \(T = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right]x\), the matrix \(A = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right]\).

So, the linear transformation matrix is \(\left[ {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right]\).