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Is this a Linear transformation

  1. Jan 20, 2008 #1
    how do i determine whether the following is a linear transformation:

    T1(x,y)=(1,y)

    i know that it must satisfy the conditions:
    (a) T(v+w)=T(v)+T(w)
    (b) T(cv)=cT(v), where c is a real constant
    and v and w are real vectors in 2D.
    v=(v1,v2) and w=(w1,w2)
    but i'm still confused.

    Thank you
     
  2. jcsd
  3. Jan 20, 2008 #2

    Ben Niehoff

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    So, using the definitions of vector addition and scalar multiplication, plug (v1, v2) and (w1, w2) into your conditions (a) and (b), and see what happens.

    (A general hint for all math stuff: Try some things and see what happens.)
     
  4. Jan 21, 2008 #3

    mathwonk

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    it should take zero to zero for one thing.
     
  5. Jan 21, 2008 #4

    HallsofIvy

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    Excellent! You just haven't DONE anything with that!

    For example, suppose you have v= [itex](x_1, y_1)[/itex] and v= [itex](x_2, y_2)[/itex]. What is T(v)? What is T(u)? What is T(v+ u)? Is T(v+u)= T(u)+ T(v)?
     
  6. Jan 21, 2008 #5
    ok i'm going to try it:
    T(v)=(1,v2)
    T(w)=(1,w2)
    T(v)+T(w)=(1, v2+w2)
    T(v+w)=T(v1+w1, v2+w2)=(1, v2+w2)
    so the first condition holds.

    i'm sure i have done something wrong
    ???
     
  7. Jan 21, 2008 #6
    Why is T(v)+T(w)=(1,v2+w2)?
    T(v)+T(w)=(1,v2)+(1,w2)=?
     
  8. Jan 21, 2008 #7
    (2,v2+w2)
    ???
     
  9. Jan 21, 2008 #8
    Ok, and is this equal to T(v+w)?
     
  10. Jan 21, 2008 #9
    T(v+w)=(v1+w1, v2+w2)
    no they're not equal......right?
    and how do i show the second condition?
     
  11. Jan 21, 2008 #10
    Same as before... experiment with a couple of values for the scalar 'c'. You'll notice right away what's going on.
     
  12. Jan 21, 2008 #11
    No, by definition of the operator T, T(v+w)=(1,v2+w2). But in either case T(v+w) is not equal to T(v)+T(w). So since it fails one of the conditions for linearity the operator is not linear.
     
  13. Jan 22, 2008 #12
    how did you get that T(v+w)=T(1, v2+w2)
    could you please show me?


    and so if i show one condition then i guess theres no need to show the other..right?
     
  14. Jan 22, 2008 #13

    mathwonk

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    to repeat myself, since T(0.v) = 0.T(v) = 0, it cannot be linear unless zero goes to zero, but T(0,0) = (1,0), which is a deal breaker.
     
  15. Jan 22, 2008 #14
    oh okay, so i guess that's enough to say that it's not linear.

    thank you.

    i was working on another question on linear transformation and i did it but i'm not sure if i did it correctly:

    T(x,y)=(y,x)

    again, let v=(v1, v2) and w=(w1,w2)

    then, T(v+w)=(v2+w2, v1+w1)

    and, T(v)+T(w)=(v2+w2, v1+w1)
    so the first condition holds.

    AND:

    let c be a constant:
    T(cv)=T(cv1,cv2)=(cv2, cv1)
    and cT(v)=c(v2, v1)=(cv2, cv1)


    is that correct?
     
  16. Jan 22, 2008 #15
    Looks fine to me... but I'd show the intermediate step in your workings of the first condition. Just as you did for the second.

    In other words, just saying "for T(v+w) we get xxxx, and for T(v)+T(w) we also get xxxx, so they must be equal" does not show how you got xxxx. Your work for the second condition, on the other hand, shows it well.
     
  17. Jan 22, 2008 #16
    i just worked on another one and i think i get the hang of it now but i just want to check this this one to make sure i know how to do it:
    T(x,y)=(x,0)

    again, let v=(v1, v2) and w=(w1,w2)

    then, T(v+w)=(v1+w1, 0)

    and, T(v)+T(w)=(v1+w1, 0)
    so the first condition holds.

    AND:

    let c be a constant:
    T(cv)=T(cv1,cv2)=(cv1, 0) = cT(v)

    is that correct?
     
  18. Jan 22, 2008 #17
    I was just doing another example and i want to check this one so i know whether i got the hang of it:

    T(x,y)=(x,0)

    again, let v=(v1, v2) and w=(w1,w2)

    then, T(v+w)=T(v1+w1, v2+w2)=(v1+w1, 0)

    and, T(v)+T(w)=(v1+w1, 0)
    so the first condition holds.

    AND:

    let c be a constant:
    T(cv)=T(cv1,cv2)=(cv1, 0)=c(v1,0)=cT(v)
    so both conditions hold

    is that correct?
     
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