Answer to Question #121520 in Linear Algebra for Tau

Question

Suppose b,c belongs to R. Define map T : R3 --> R2 by
T.(x,y,z)= (2x - 4y + 3z +b, 6xcxyz)
Show that T is linear if and only if b = c = 0.

Answer 1

$T.(x,y,z)= (2x - 4y + 3z +b, 6xcxyz)$

For a mapping to be linear,

1. $T(\alpha x,\alpha y,\alpha z)=\alpha (T(x,y,z))$ $\forall\ \alpha \in \ R$
2. $T(x_1+x_2,y_1+y_2,z_1+z_2)=T(x_1,y_1,z_1)+T(x_2,y_2,z_2)$ $\forall\ \alpha \in \ R$

So, if $T$ is linear,then

$T(\alpha x,\alpha y,\alpha z)=$ $(2\alpha x - 4\alpha y + 3\alpha z +b, 6(\alpha x)c(\alpha x)(\alpha y)(\alpha z))=$ $(2\alpha x - 4\alpha y + 3\alpha z +b,$ $6(\alpha^4 )cx^2yz)$

This must be equal to $(\alpha (2x - 4y + 3z +b), \alpha (6xcxyz))$ .

So,$\alpha (2x - 4y + 3z +b)$ $=(2\alpha x - 4\alpha y + 3\alpha z +b)$

$\implies b\alpha = b\implies b=0$

And,$\alpha (6xcxyz))$ $=$ $6(\alpha^4 )cx^2yz)\implies$ $6\alpha x^2yzc(1-\alpha ^3)=0$

And $\alpha ,x,y,z$ can have any values.

So,for the above statement to be true,$c=0.$