Question

In this exercise, T: R2 → R2 is a function. For each of the following parts, state why T is not linear. (a) T(a1, a2)= (1, a2) (b) T(a, a2) (a1,ai) (c) T(a1,a2)= (sin a1,0) (d) T(al, aa) = (lal, a2) (e) T(a1, d12) = (a1 + 1, a2)

Answer 1

Answer:

• a) T is not homogeneous
• b) T is not additive
• c) T is not homogeneous
• d) T is not additive
• e) T is not additive

Step-by-step explanation:

In each case there is an example where one property of linear functions fails.

• a) $T(2(1,0))=T((2,0))=(1,0); 2T((1,0))=2(1,0)=(2,0)$. These vectors are not equal, then T doesn't satisfy the condition of scalar multiplication (homogeneity).

• b) $T((1,2)+(2,3))=T(3,5)=(3,9); T((1,2))+T((2,3))=(1,1)+(2,4)=(3,5)$. Because these vectors are not equal, T doesn't satisfy the property of vector addition (additivity).

• c) $T(\frac{1}{2}(\pi,0))=T((\frac{\pi}{2},0))=(\sin(\frac{\pi}{2}),0)=(1,0); \frac{1}{2}T((\pi,0))=\frac{1}{2}(0,0)=(0,0)$ so T is not homogeneous.

• d) $T((-1,0)+(1,0))=T(0,0)=(0,0); T((-1,0))+T((1,0))=(1,0)+(1,0)=(2,0)$ then T is not additive.

• e) $T((0,0)+(1,0))=T(1,0)=(2,0); T((0,0))+T((1,0))=(1,0)+(2,0)=(3,0)$ then T is not additive.