# Which of the following are Linear Transformations?

1. Apr 29, 2010

### newtomath

Below is a HW problem which I believe is correct. Can you guys take a look and advise?

Which of the following are linear transformations
A) L(x,y,z)= (0,0)
B) L(x,y,z)= = (1,2,-1)
C) L(x,y,z)= ( x^2 +y , y-z)

To prove these relationships are linear transformations, they must satisfy the below conditions:

1) L(u+v)= L(u) + L(v) for all u and v in the space.
2) L(ku) = k*L(u) for all u in the space for every k scalars.

we can use u= (u1, u2, u3) and v=( v1, v2, v3)

A) From the problem we know : L(u)= (0,0) and L(v)= (0,0). L(u+v)= L (u1+ v1, u2 +v2, u3 +v3) = (0,0). L(u)+L(v) = (0,0) +(0,0) = (0,0), which is what we are looking for. so this point is satisfied.

L(ku) = k*L(u). L(ku)= k*(0,0) = (k*0, k*0) = (0,0). So this point is satisfied as well. A is a linear transformation.

B) Given u and v from above, we have L(u)= (1,2,-1) and L(v)= (1,2,-1). L(u+v)= L (u1+ v1, u2 +v2, u3 +v3) = (1,2,-1). L(u)= (1,2,-1) and L(v)= (1,2,-1) so L(u)+L(v) = (1,2,-1) +(1,2,-1) = (2,4,-2). This is inconsistent from our goal of (1,2,-1) so this is not a linear transformation. the second condition does not need to be proved since the first condition has not been met.

B is not a transformation

C) We have L(u)= (u1^2 + u2, u2-u3 ) and L(v)= (v1^2 + v2, v2-v3 ). L(u+v)= L (u1+ v1, u2 +v2, u3 +v3) = { ((u1+v1)^2+ (u2+v2)) , ((u2+v2) - (u3+v3)) }
L(u)+L(v) = (u1^2 + u2, u2-u3 ) + (v1^2 + v2, v2-v3 ) = { (u1^2 + u2 + v1^2 + v2 ) , u2-u3 + v2-v3) }

C is not a transformation

2. Apr 30, 2010

Right.