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Homework Help: Linear Relations :?

  1. Oct 2, 2005 #1
    "Determine whether each of the following functions is a linear transformation. If you think the
    function is a linear transformation then prove that it is. If you think the function is not a linear
    transformation then explain why.
    (a) T : R2 ! R2; T(x, y) = (x − 2y, 2xy)."

    I don't want an answer, i won't to learn how to do these types of problems, i never understood it at all, :confused: And stuff like this ??

    T : R4 ! R3; T(x, y, z,w) = (x − 2y, y + w, z).
    Find a matrix A such that T(u) = Au, where u 2 R4.

    Consider the following linear transformations:
    T : R2 ! R3; T(x, y) = (x + 3y, x − y, 2x),
    S : R3 ! R3; S(x, y, z) = (x − 2y, y + z,−z).
    (a) Is the composite linear transformation S  T defined? If S  T is defined, write down
    formula for S  T.
    (b) Does the linear transformation S have an inverse? Give a reason for your answer.

    Can you guys give me a link or something where i can learn wtf this means i never understood linear transformations, and i bombed out in the exam, but i want to get better. Thanks, BTW I DONT WANT ANSWERS :yuck: Just want to know how to do all kinds of problems like above
  2. jcsd
  3. Oct 2, 2005 #2
    To do these problems, you need to know only one thing, the definition of a linear transformation, which should be in your text. Note that an immediate theorem from the definition is that a linear transformation on a vector space is completely defined by its action on the basis vectors of a basis of that space. This is where you can get your matrix components from.
  4. Oct 2, 2005 #3


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    Science Advisor

    Since you want to "know how to do these types of problems", the answer is "Use the Definitions". To show that something is a "whatever", show that it satisfies the conditions for "whatever". As hypermorphism said, a linear transformation, L, satisfies
    L(au+ bv)= aLu+ bLv. That's the definition (some times given as two parts: L(au)= aLu and L(u+ v)= L(u)+ L(v) but it's the same thing). Since that is an equation, plug what you are given into the equation and see if it works!

    For example, suppose T is defined by T(x,y)= (2x+y, x2)
    Set u= (p,q), v= (r,s) )(p,q,r,s are numbers, of course). Then Tu= T(p,q)= (2p+q,p2), Tv= (2r+s,s2), aTu= (a(2p+q),ap2)= (2ap+aq,ap2), and bTv= (2br+ bs,bs2). au= (ap,aq) and bv= (br,bs) so
    T(au+ bv)= T(ap+br,aq+bs)= (2(ap+br)+aq+bs,(aq+bs)2)= ((2ap+aq)+(2br+bs),a2q2+2abqs+ b2s2). The first component of that is, in fact, the same as aTu+ bTv but the second is NOT- it has that extra 2abqs in it. T is not "linear".
    In fact, although it is not a proof, observing that "squaring" is NOT linear itself should have told you that this would happen. If you happened to notice that this was not linear,you could have done this by giving a "
  5. Oct 2, 2005 #4
    Thank you so much, that helped me :cool: Appreciate the help dude
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