Answer to Question #270204 in Linear Algebra for Gestavo

Question #270204

Determine if the following transformation is a linear transformation:

T (x1, x2) → (cos(x1), sin(x2))

T = R2→R2



1
Expert's answer
2021-11-23T11:50:20-0500

Given T(x1x2)=(cosx1sinx2)Let α be a scalar and (y1y2) be vector then: T((x1x2)+α(y1y2))=T(x1+αy1x2+αy2)=(cos(x1+αy1)sin(x2+αy2))=(cosx1cosαy1sinx1sinαy1sinx1cosαy1+cosx1sinαy1)And T(x1x2)+αT(y1y2)=(cosx1sinx2)+α(cosy1siny2)=(cosx1+αcosy1sinx2+αsiny2)(cosx1cosαy1sinx1sinαy1sinx1cosαy1+cosx1sinαy1)=T((x1x2)+α(y1y2))    T(x1x2)+αT(y1y2)T((x1x2)+α(y1y2))The given transformation is not a linear transformation.\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.}


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