Determine if the transformation is linear if T(x, y)= (x+1, 2y)

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Homework Statement



Determine if the transformation T: [tex]R^2\rightarrow R^2[/tex] is linear if T(x, y)= (x+1, 2y)


Homework Equations


1. T(u + v) = T(u) + T(v)
2. T(c*u) = cT(u)
3. T(0) = 0

The Attempt at a Solution



(1). I'm not sure how to prove the first condition (additivity). Can anyone help?

(2). T(c x,c y) = (c x+1, c 2y) = c(x+1, 2y) =c T(x,y)

For some scalar c.

(3). In (2) if c=0
T(0 x,0 y) = (0 x+1, 0 2y) = 0(x+1, 2y) =0
 

Answers and Replies

  • #2
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Homework Statement



Determine if the transformation T: [tex]R^2\rightarrow R^2[/tex] is linear if T(x, y)= (x+1, 2y)


Homework Equations


1. T(u + v) = T(u) + T(v)
2. T(c*u) = cT(u)
3. T(0) = 0

The Attempt at a Solution



(1). I'm not sure how to prove the first condition (additivity). Can anyone help?
You may or may not be able to prove any of these conditions. What you need to do is to verify that they are all true, or show that one or more are not true.

For the first one, let u = (u1, u2) and v = (v1, v2)
Is T(u + v) = T(u) + T(v)?
(2). T(c x,c y) = (c x+1, c 2y) = c(x+1, 2y) =c T(x,y)
In the previous equation, you are trying to show that T(cu) = cT(u), for any scalar c, and any vector u.

If u = (x, y), as you are assuming, what is T(cx, cy)? What you have after that step is not correct.
For some scalar c.

(3). In (2) if c=0
T(0 x,0 y) = (0 x+1, 0 2y) = 0(x+1, 2y) =0
No, T(0) is not 0. Remember that here 0 is a vector - (0, 0). What does T do to (0, 0)?
 
  • #3
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So, for the first condition, T(u+v) = (u1 + v1+1, 2 u2+v2) is not equal to (u1, 2u2) + (v1+1, 2u1) = T(u)+T(v).

For the third condition, yes it will be (1,0) noy (0,0).

If u = (x, y), as you are assuming, what is T(cx, cy)? What you have after that step is not correct.

T(cx,cy)=(cx+1,2cy) = c(x+1,2y)= cT(x,y)

Is it right?
 
Last edited:
  • #4
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So what do you conclude about this transformation?
 
  • #5
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It is not linear since it violates at least one of the conditions for linearity. Thanks!!
 

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