Question

Determine if the following transformation is a linear transformation:

T (x₁, x₂) → (cos(x₁), sin(x₂))

T = R²→R²

Answer 1

$\displaystyle\text{Given } T\binom{x_1}{x_2} = \binom{\cos x_1}{\sin x_2} \\ \text{Let $\alpha$ be a scalar and }\binom{y_1}{y_2} \text{ be vector then: } \\ T\left(\binom{x_1 }{x_2} + \alpha \binom{y_1}{y_2}\right) = T\binom{x_1+ \alpha y_1}{x_2 + \alpha y_2} \\ \qquad \qquad \qquad \qquad\quad= \binom{\cos(x_1 + \alpha y_1)}{\sin(x_2+ \alpha y_2)} = \binom{\cos x_1\cos \alpha y_1 - \sin x_1\sin \alpha y_1}{\sin x_1\cos \alpha y_1 + \cos x_1\sin \alpha y_1} \\ \text{And } \\ T\binom{x_1 }{x_2} + \alpha\, T\binom{y_1}{y_2} = \binom{\cos x_1}{\sin x_2} + \alpha \, \binom{\cos y_1}{\sin y_2} = \binom{\cos x_1 +\alpha \cos y_1}{\sin x_2 + \alpha \sin y_2} \ne \binom{\cos x_1\cos \alpha y_1 - \sin x_1\sin \alpha y_1}{\sin x_1\cos \alpha y_1 + \cos x_1\sin \alpha y_1} = T\left(\binom{x_1 }{x_2} + \alpha \binom{y_1}{y_2}\right) \\ \implies T\binom{x_1 }{x_2} + \alpha\, T\binom{y_1}{y_2} \ne T\left(\binom{x_1 }{x_2} + \alpha \binom{y_1}{y_2}\right) \\ \therefore \text{The given transformation is not a linear transformation.}$