Question

Suppose b,c element of R, and T: R³ - R² dened as T (x;y;z) = (2x4y+3z+b,6x+cxy): Show that T is linear if and only if b = c = 0.

Answer 1

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

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

$$T(x_1,y_1,z_1) + T(x_2,y_2,z_2)$$

 .

$$T(x_1+x_2,y_1+y_2,z_1+z_2)$$

$$= \big(2(x_1+x_2)-4(y_1+y_2) + 3(z_1+z_2)+b,$$

$$6(x_1+x_2)+c(x_1+x_2)(y_1+y_2)\big).$$

$$T(x_1,y_1,z_1) + T(x_2,y_2,z_2)$$

$$= (2x_1-4y_1+3z_1+b,6x_1+cx_1y_1)$$

$$+ (2x_2-4y_2+3z_2+b,6x_2+cx_2y_2 )$$

$$= \big(2(x_1+x_2)-4(y_1+y_2) + 3(z_1+z_2)+2b,$$

$$6(x_1+x_2)+c(x_1y_1+x_2y_2)\big).$$

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

$$6(x_1+x_2)+c(x_1+x_2)(y_1+y_2)$$

 

$$=6(x_1+x_2)+c(x_1y_1+x_2y_2)$$

$$c(x_1+x_2)(y_1+y_2)=c(x_1y_1+x_2y_2)$$

$$c(x_1y_2+x_2y_1)=c$$

for every $x_1,x_2,y_1,y_2,z_1,z_2\in\R.$ 

Let, for example, $x_1=y_1=x_2=y_2 = 1,$  then

$$2c=c=>c = 0.$$

 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.

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

For every $x_1,x_2,y_1,y_2,z_1,z_2\in\R \; \;$ 

$$T(x_1+x_2,y_1+y_2,z_1+z_2)$$

$$= (2(x_1+x_2)-4(y_1+y_2)+3(z_1+z_2),6(x_1+x_2))$$

$$= (2x_1-4y_1+3z_1,6x_1) + (2x_2-4y_2+3z_2,6x_2)$$

$$= T(x_1,y_1,z_1) + T(x_2,y_2,z_2),$$

so $T$ is linear.