SOLUTION: is the following a linear transformation from R^3 to R^2?
L(x1,x2,x3) = (x2-x3,x1+2) ?
so i know how to prove that this is not closed under scalar multiplication and addition
Question 18994: is the following a linear transformation from R^3 to R^2?
L(x1,x2,x3) = (x2-x3,x1+2) ?
so i know how to prove that this is not closed under scalar multiplication and addition (due to the "2" in the second entry). so then is my answer just "no, it is not a linear transformation from R^3 to R^2 because it is not closed under scalar multiplication." ?
what about linear transformation from R^3 to R^4: L(x1,x2,x3) = (x2-x3,x1*x2,2x2-x1,x3+x2) ?
for this one, i dont quite understand how to go about it because the second entry is a product of two numbers and not just a sum. so then whould it count as being closed under scalar multiplication and addition? plus i remember seeing somewhere if R^m ---> R^n, then m must be greater than or equal to n. so then by this rule, the answer is automatically no because we are going from R^3 to R^4 and (obviously) 4 is greater than 3. i'm just not sure about all of this...
thank you so much for your help : ) i greatly appreciate it!!! Answer by khwang(438) (Show Source): You can put this solution on YOUR website! a linear transformation from to ?
L(x1,x2,x3) = (x2-x3,x1+2) ?
(i) It is still closed under scalar multiplication and addition but
not linear. Simly, because L(0,0,0) != (0,0).
Since, for any a linear transformation L, L(0) = L(0+0)= L(0) + L(0)
so L(0) = 0 for zero vector.
In general, such L to if and only if each n component
of L is a linear function. Now the 2nd component x1+2 is not linear.
[You idea was Ok with wrong reason.]
so i know how to prove that this is not closed under scalar multiplication and addition (due to the "2" in the second entry). so then is my answer just "no, it is not a linear transformation from R^3 to R^2 because it is not closed
under scalar multiplication." ?
NO !!!!
what about linear transformation from R^3 to R^4: L(x1,x2,x3) = (x2-x3,x1*x2,2x2-x1,x3+x2) ?
No, since the 2nd component x1*x2 is not linear.
More precisely,
L(x1,x2,x3) = (x2-x3,x1*x2,2x2-x1,x3+x2)
Since (1,1,0) = (1,0,0) +(0,1,0).
L(1,1,0) = (1, 1,1, 1)
L(1,0,0) = (0, 0, -1, 0) and L(0,1,0) = (1, 0, 2, 1)
but L(1,0,0) + L(0,1,0) = (1,0,1,1) != L(1,1,0)
Hence,L is not a linear transformation.
Another wrong idea:
" if ---> , then m must be greater than or equal to n. "
No, m ,n can be any two non-negative integers, not restricting to m<=n.
Good luck !
Kenny