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Determination of Linear transformation

  • Thread starter Maxwhale
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  • #1
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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, [tex]\sqrt[3]{xy}[/tex] )

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([tex]\sqrt[3]{xy}[/tex] )
T(ru ) = r(x + y2 , [tex]\sqrt[3]{xy}[/tex] )

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 !
 

Answers and Replies

  • #2
Redbelly98
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If u=(x,y), what is ru?
 
  • #3
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(rx, ry)
 
  • #4
Redbelly98
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Yes.

So T(ru) = T(rx, ry) = ???
 
  • #5
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yeah i have done that
 
  • #6
Redbelly98
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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, [tex]\sqrt[3]{xy}[/tex] )

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

The Attempt at a Solution


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

so it suffices the first condition, right?
 
Last edited:
  • #7
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should it be

let u = (x1, x2);
T(ru ) = T(rx1, rx2)
T(ru )= ( (rx1 + (rx2)2), [tex]\sqrt[3]{(rx)(ry)}[/tex] )
 
  • #8
Redbelly98
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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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