Question

How do you show whether T(x,y)=(cube root(x), cube root(y)) is linear? I know we need to show if has additivity and homogeneity

Answer 1

if T is linear then

T((2,2))=T((1,1)+(1,1))=T((1,1))+T((1,1))

is this true?

Answer 2

I would say no b/c (1,1) doesn't necessary equal to T((1,1))

Answer 3

But how does relate to cube root?

Answer 4

\[T(2,2)=(\sqrt[3]{2},\sqrt[3]{2})\neq (2,2)=(\sqrt[3]{1},\sqrt[3]{1})+(\sqrt[3]{1},\sqrt[3]{1})=T(1,1)+T(1,1)\]

therefore is is not Linear

Answer 5

Yeah that breaks the homogeneity rule right, how would you show it for the additivity rule?

Answer 6

It's suppose to be T(u) + T(v) = T(u+v)

Answer 7

I just don't get what is v and what is u?

Answer 8

if linear then
T(u) + T(v) = T(u+v)

and

T(cv)=cT(v)

Answer 9

v and u are vectors in your space...in this case $$R^2$$

Answer 10

your vectors are 2-tuples

Answer 11

another way to show that it is not linear

$$T(2,2)=(\sqrt[3]{2},\sqrt[3]{2})$$

but if T is linear we should have

$$T(2,2)=T(2(1,1))=2T(1,1)=2(\sqrt[3]{1},\sqrt[3]{1})=2(1,1)=(2,2)$$

these two are not the same

Answer 12

T(cx, cy)=(cube root(cx), cute root(cy) = cube root c(cube root x, cube root y)= (cube root c)T(x,y)

Answer 13

That's how my prove went for homogeneity I don't get where you got your numbers from

Answer 14

(cube root c)T(x,y) does not equal T(cx, cy)

Answer 15

to disprove something all we need is a counter example. I provided a counter example.

Answer 16

If T was linear my example should have worked...they didn't . That proves T was not linear