Question

Someone please helpppIs L(x,y)=(x+y,x-y) a linear transformation?(Show with full working)

Answer 1

if L(u + v) = L(u) + L(v) ,
and
L ( c*u) = c* L(u), then its a linear transformation

Answer 2

and the definition says a function from R^n to R^m, here n=m = 2

Answer 3

Yup i know the conditions but i dont know how to apply it.

Answer 4

Could you help me solve in terms of two vectors. For both right side and left side

Answer 5

L(x,y)=(x+y,x-y)

Let
u = (u1, u2)
v = (v1,v2)
Then
u + v = ( u1 + v1 , u2 + v2)

Therefore
L ( u + v ) = L ( u1+v1, u2 + v2) = ( u1 + v1 + u2 + v2 , u1 + v1 - ( u2 + v2) )

Answer 6

okay is this the final answer ?? or we have to check scalar multiplication too?

Answer 7

now its not clear how to proceed, lets start from the other side

evaluate

L(u) + L(v)

Answer 8

yes please im watching

Answer 9

u there??

Answer 10

u = (u1,u2)
v = (v1,v2)

L( u ) = ( u1 + u2, u1 - u2 )
L( v) = (v1 + v2 , v1 - v2)


L(u) + L(v)
= ( u1 + u2, u1 - u2 ) + (v1 + v2 , v1 - v2)
= ( u1 + u2 + v1 + v2 , u1 - u2 + v1 - v2)
= ( u1 + v1 + u2 + v2, u1 + v1 - (u2 + v2) )

Answer 11

now the last expression, that is what we got for L ( u + v )

Answer 12

okayy...now its much clear

Answer 13

do we need to chk multiplication also

Answer 14

thanks mate

Answer 15

hey, how about this

Answer 16

you know vectors are bold, but vector components are normal shaded

Answer 17

yup

Answer 18

i cant draw bold vectors, so i will use capital letters

Answer 19

ok

Answer 20

Based on perl's definitions, we can say

L ( u + v ) = L ( u1+v1, u2 + v2)
L ( u + v ) = ( u1 + v1 + u2 + v2 , u1 + v1 - ( u2 + v2) )
L ( u + v ) = ( u1 + v1 + u2 + v2 , u1 + v1 - u2 - v2) )
L ( u + v ) = ( (u1 + u2) + (v1 + v2) , (u1 - u2) + (v1 - v2)) )
L ( u + v ) = (u1 + u2,u1 - u2) + (v1 + v2,v1 - v2)
L ( u + v ) = L(u) + L(v)

So that takes care of the first part. You need to do the same for L(c*u) = c*L(u)

Answer 21

Given
L(x,y)=(x+y,x-y)
show L is a linear transformation from R^2 to R^2

Let
U = (u1, u2)
V = (v1,v2)

Then
U + V = ( u1 + v1 , u2 + v2)

Therefore

L(U) + L(V)
= ( u1 + u2, u1 - u2 ) + (v1 + v2 , v1 - v2)
= ( u1 + u2 + v1 + v2 , u1 - u2 + v1 - v2)
= ( u1 + v1 + u2 + v2, u1 + v1 - (u2 + v2) )
= L(u1 + v1, u2 + v2)
= L ( U + V )

Answer 22

ohhhhhhhhhhhh i get that thanks jim n perl. you rock

Answer 23

L( c*U) = ( c*u1 + c*u2 , c*u1 - c*u2) = c * ( u1 + u2 , u1 - u2) = c* L( U )

Answer 24

thanks a million