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Linear algebra: transformations

  1. Oct 26, 2006 #1
    Hello, I'll be online until I get this one completely figured out, so baby steps are for the win here.

    Let L1:U->V and L2:U->W be linear transformations, and let L = L2 * L1 be the mapping defined by:

    L(u) = L2(L1(u))

    for each u which lies in U. Show that L is a linear transformation mapping U into W.

    So basically, should I first show that L1(u) is a valid linear transform?, and then show that L2, is, too?
    Last edited: Oct 26, 2006
  2. jcsd
  3. Oct 26, 2006 #2


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    Hm, I believe it's enough to assume L1 and L2 are linear transformations, and then proove that L is one, too.
  4. Oct 26, 2006 #3


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    Is L2 from U to W or from V to W?
  5. Oct 26, 2006 #4
    V to W! thank you for that!
    What about this? (in progress (latex noobie))

    [tex]L(u) = L_2 (L_1 (u) )}[/tex]

    [tex]L(u_1 + u_2) = L_2 (L_1 (u_1 + u_2) )}[/tex]

    [tex]L(\alpha u_1 + \beta u_2) = L_2 (L_1 (\alpha u_1 + \beta u_2) )}[/tex]

    [tex]L(\alpha u_1 + \beta u_2) = L_2 (\alpha L_1(u_1) + \beta L_1 (u_2) )}[/tex]

    [tex]L(\alpha u_1 + \beta u_2) = \alpha L_2 (L_1(u_1)) + \beta L_2 (L_1(u_2)) [/tex]

    [tex]\alpha L(u_1)+ \beta L(u_2) = \alpha L_2 (L_1(u_1)) + \beta L_2 (L_1(u_2)) [/tex]

    Is this on the right track? If so, should I break it up into two pieces, and show that L1(u) is surely a mapping into V, and then show that L2(v) is surely a mapping into W?

    Or am I way off
    Last edited: Oct 26, 2006
  6. Oct 27, 2006 #5


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    That's exactly right. Now all you have to do is note that
    [tex]L_2(L_1(u_1))= L(u_1)[/tex]
    [tex]L_2(L_1(u_2))= L(u_2)[/tex]
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