Answer to Question #121752 in Linear Algebra for Henry

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

Let us consider $T(x_1+x_2,y_1+y_2,z_1+z_2).$ It should be equal to $T(x_1,y_1,z_1) + T(x_2,y_2,z_2)$ .

$T(x_1+x_2,y_1+y_2,z_1+z_2) = \big(2(x_1+x_2)-4(y_1+y_2) + 3(z_1+z_2)+b, 6c(x_1+x_2)^2(y_1+y_2)(z_1+z_2)\big).$

$T(x_1,y_1,z_1) + T(x_2,y_2,z_2) = (2x_1-4y_1+3z_1+b,6cx_1^2y_1z_1) + (2x_2-4y_2+3z_2+b,6cx_2^2y_2z_2 ) = \big(2(x_1+x_2)-4(y_1+y_2) + 3(z_1+z_2)+2b, 6c(x_1^2y_1z_1+x_2^2y_2z_2)\big).$

Therefore, $2(x_1+x_2)-4(y_1+y_2) + 3(z_1+z_2)+b = 2(x_1+x_2)-4(y_1+y_2) + 3(z_1+z_2)+2b$ for every $x_1,x_2,y_1,y_2,z_1,z_2\in\R.$ It is true if and only if $2b=b \; \Rightarrow b = 0.$

Also it should be true that $6c(x_1+x_2)^2(y_1+y_2)(z_1+z_2) = 6c(x_1^2y_1z_1+x_2^2y_2z_2)$ for every $x_1,x_2,y_1,y_2,z_1,z_2\in\R.$ Let, for example, $x_1=y_1=z_1 = 1, \; x_2=y_1=z_1=2$ , therefore

$6c\cdot81= 6c\cdot65 \Rightarrow c = 0.$ So $c$ can be only equal to 0.

So we conclude that if T is linear, b=0 and c = 0. Now we should prove that for $b=c=0$ T is linear.

$T(x,y,z) = (2x - 4y + 3z, 0)$ . For every $x_1,x_2,y_1,y_2,z_1,z_2\in\R \; \;$ $T(x_1+x_2,y_1+y_2,z_1+z_2) = (2(x_1+x_2)-4(y_1+y_2)+3(z_1+z_2),0) = (2x_1-4y_1+3z_1,0) + (2x_2-4y_2+3z_2,0) = T(x_1,y_1,z_1) + T(x_2,y_2,z_2),$

so T is linear.