Determination of Linear transformation

Maxwhale
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Homework Statement



Determine if the following T is linear tranformation, and give the domain and range of T:

T(x,y) = (x + y2, \sqrt[3]{xy} )

Homework Equations



T ( u + v) = T(u) + T(v)

T(ru) = rT(u)



The Attempt at a Solution


1)
let u = (x1, x2);
T(ru ) = T(rx1, rx2)
T(ru )= r(x + y2) , r(\sqrt[3]{xy} )
T(ru ) = r(x + y2 , \sqrt[3]{xy} )

so it suffices the first condition, right?

2)
let u = (x1, y1) and let v = (y1, y2);
T ( u + v ) = T ( x1 + y1, x2 + y2)
T ( u + v ) = Here I am confused with the term ( x + y2)
T ( u + v )


Any help please !
 
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If u=(x,y), what is ru?
 
(rx, ry)
 
Yes.

So T(ru) = T(rx, ry) = ?
 
yeah i have done that
 
Okay, but you did it wrong.

What do you get when you substitute rx for x, and ry for y, into
T(x,y) = (x + y2, \sqrt[3]{xy} )

EDIT: FYI this is the part that I'm trying to correct:
Maxwhale said:

The Attempt at a Solution


1)
let u = (x1, x2);
T(ru ) = T(rx1, rx2)
T(ru )= r(x + y2) , r(\sqrt[3]{xy} )
T(ru ) = r(x + y2 , \sqrt[3]{xy} )

so it suffices the first condition, right?
 
Last edited:
should it be

let u = (x1, x2);
T(ru ) = T(rx1, rx2)
T(ru )= ( (rx1 + (rx2)2), \sqrt[3]{(rx)(ry)} )
 
Yes. Although x1 is x, and x2 is y, of course.

Next (as you know), you compare that expression with rT(u).
 
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