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Linear Functionals Questions.

  1. Jun 3, 2006 #1
    1) let S:U->V T:V->W be linear operators, show that: (ToS)^t=S^toT^t.
    2) let T:V->U be linear and u belongs to U, show that u belongs to Im(T) or that there exist [tex]\phi\inV*[/tex] such that [tex]T^{t}(\phi)=0[/tex] and [tex]\phi(u)=1[/tex]

    about the first question here what i tried to do:
    [tex](ToS)^{t}(\phi(v))=\phi o(ToS)(v)=\phi(T(S(v))=T^{t}(\phi(S(v))=T^{t}(S^{t}(\phi(v))=T^{t}oS^{t}[/tex]

    about the second question the first part (of u belongs to ImT) i did it (simply the defintion of ImT), but i don't know how to to approach the latter.
    ofcourse i have [tex]T^{t}(\phi(v))=\phi(T(v))=\phi(u)[/tex] where v belongs to V, but other than this i dont know how to proceed.
     
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  3. Jun 3, 2006 #2

    matt grime

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    what does any of this have to do with linear functionals?

    1 is clear from the definitions, 2 has some latex graphics that won't render properly for me.
     
  4. Jun 4, 2006 #3

    AKG

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    No, that's not what it means. It wants to prove that for any u in U, either u is in im(T) or there exists [itex]\phi \in U^*[/itex] (you need to put a space after the \in and before the V* in your LaTex, and also [itex]\phi[/itex] needs to be in U*, not V*) such that [itex]T^t(\phi ) = 0[/itex] and [itex]\phi (u) = 1[/itex]. It's not like for any u in U, u is in im(T) by definition. im(T) is not equal to U, im(T) is a subset of U. If im(T) happens to be all of U, then T is called "surjective", as I'm sure you know.

    In both 1 and 2, your notation is a little off. Where you have:

    [tex]T^t(\phi (v))[/tex]

    that doesn't really make sense, it should be:

    [tex][T^t(\phi )](v)[/tex]

    Similarly,

    [tex](T \circ S)^t (\phi (v))[/tex]

    doesn't make sense, you need:

    [tex][(T \circ S)^t (\phi )](v)[/tex]

    I would go over your proof for the first part again. The idea is correct, but you make problems like the one pointed out above in two other places.

    Getting back to the second problem, it asks that [itex]\phi (u) = 0[/itex] so it's clear that [itex]\phi[/itex] ought to be in U*. Since T is a mapping from V to U, then Tt is a mapping from U* to V*. The condition [itex]T^t (\phi ) = 0[/itex] means that [itex]T^t (\phi )[/itex] needs to be the zero functional in V* (since Tt maps things from U* (like [itex]\phi[/itex]) to things in V*). That means that for all v in V, we want:

    [tex][T^t(\phi )](v) = 0[/tex]

    which by definition means that for all v in V, we want:

    [tex]\phi (T(v)) = 0[/tex]

    In other words, we want [itex]\phi (im(T)) = \{0\}[/itex] but of course, we also want [itex]\phi (u) = 1[/itex]. Well if u is in im(T), then it is impossible to meet these conditions. But if u is not in im(T), then it is possible to meet these conditions. You are asked to prove that it is indeed possible to meet this conditions when u is not in im(T).
     
  5. Jun 4, 2006 #4
    about V*, it's not my mistkae, this is how it was written in my text.

    about my first question, [tex][(ToS)^{t}(\phi)](v)=\phi(ToS)(v)=\phi(T(S(v))=T^{t}(\phi(S(v))=T^{t}(S^{t}(\phi(v)))=T^{t}oS^{t}(\phi)(v)[/tex]
    which as you can see it's not what i needed.
    anyway, about my second question, i know that Ker(T)={0} but i don't think that the same applies for Im(T) unless im given that T^t(phi)(v)=0.
     
  6. Jun 4, 2006 #5

    matt grime

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    Well, you have a mistake immediately after the 4th equals sign. You need to remember that you're applying

    [tex]T^t\phi[/tex]

    to the vector Sv. If it helps write [itex]\psi[/itex] for [itex]T^t\phi[/itex], what happens if you apply [itex]\psi[/itex] to Sv? You get [itex]S^t\psi v[/itex]


    For the second one, how do you know ker{T}=0? It doesn't say that T is injecitive in your post.

    You construct linear functionals be defining them on a basis and extending them linearly to all of the space.
     
  7. Jun 4, 2006 #6
    thanks for the first question.
    about the second i know that the definition of Ker(T) is Ker(T)={[tex]v \in V[/tex] |T(v)=0} so if we set T(v)=v then Ker(T)={0}.

    about what AKG typed:
    if u isn't in Im(T) then u isn't in U and T(v) !=u for v in V, but how does this bit of information helps to prove that T^t(phi)=0 and [tex]\phi(u)=1[/tex]?
     
  8. Jun 4, 2006 #7

    AKG

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    You still haven't written the first one correctly, i.e. as is, it doesn't make sense. As for the second, why are you looking at Ker(T)? And:

    if u isn't in Im(T) then u isn't in U

    is absurd. And you're not trying to prove that Tt(phi) = 0 and phi(u) = 1, that wouldn't even make sense. What is phi? You're trying to prove that THERE EXISTS a phi such that Tt(phi) = 0 and phi(u) = 1.

    Do you understand what the question is asking?
     
  9. Jun 4, 2006 #8
    the defintion of Im(T) is: Im(T)={[tex]v \in V[/tex]|T(v)=u, [tex]u \in U[/tex]}. from this you can see that if v doesnt belong to ImT then it doesnt belong to V.
    anyway, you said that from u doesnt belong to ImT I should deduce that T^t(phi)=0 and phi(u)=1, i don't know how to do so.
     
  10. Jun 4, 2006 #9
    i can prove that:
    [tex]T^{t}(\phi-\phi)=T^{t}(\phi)-T^{t}(\phi)=0[/tex] but i dont think this is what needed to be proven, cause it's quite trivial.
     
  11. Jun 4, 2006 #10

    AKG

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    That is not the definition of Im(T).
    No I didn't. I said that if u doesn't belong to Im(T), then you can prove that THERE EXISTS phi in U* such that Tt(phi) = 0 and phi(u) = 1.
     
  12. Jun 4, 2006 #11
    AKG,then what is the definition of Im(T). (the definition ive given is from my textbook).

    so how from u not in Im(T) I can deduce that there exist T^t(phi)=0 and phi(u)=1. (i have a feeling that you don't want to help anyway, because i was the one who asked in the first place the question, so i guess even if i didnt in my last post say "there exists" this should be obvious to you that that was what i meant).
     
  13. Jun 4, 2006 #12

    matt grime

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    I don't see why you would want to look at the identity map. What has that got to do with the question?

    that is not what AKG wrote.
     
  14. Jun 4, 2006 #13

    matt grime

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    That is not the definition from your textbook. Try looking at it again. The image of T is the set of vectors T maps onto. In your definition v was to be in the Image and then you applied T to v, this does not in general make sense.
     
  15. Jun 4, 2006 #14

    AKG

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    What book are you using? I might suggest getting another one. Anyways, you've asked "so how from u not in Im(T) I can deduce that there exist T^t(phi)=0 and phi(u)=1" but I'm not going to just give you the answer. If you can understand post#3, then you should be able to figure it out.

    Just to see where are, how would you answer this question: Prove that there exists a function f in (R4)* such that f(1,0,0,0) = 5 and f(0,1,0,0) = 2.
     
  16. Jun 4, 2006 #15

    matt grime

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    Let's do a toy example. Suppose that T is a map from R^3 to R^3 and it sends (a,b,c) to (a,b,0). So any of the vectors (0,0,c) is not in the image. Can you see how to do the question here?


    Just think about finite dimensional vector spaces with, say, column vectors of lenght n. The dual space, the space of linear functionals is just the set of row vectors and composition of a functional and a vecor is the ordinary product of matrices. Linear maps are just matrices. In these familiar terms it is very easy to see how to compose non-zero matrices to get a zero matrix.
     
  17. Jun 4, 2006 #16
    **** i do have a mistake, it should be:
    Im(T)={[tex]u \in U[/tex]|T(v)=u [tex]v \in V[/tex]}

    anyway, to you question:"Just to see where are, how would you answer this question: Prove that there exists a function f in (R4)* such that f(1,0,0,0) = 5 and f(0,1,0,0) = 2."
    suppose f is in (R4)*,f(x,y,z,w)=f(x,0,0,0)+f(0,y,0,0)+f(0,0,z,0)+f(0,0,0w), if we set: f(x,0,0,0)=5x f(0,y,0,0)=2y f(0,0,z,0)+f(0,0,0,w)=0 then f is a linear functional.
    am i wrong, by the way how is this got to do with my question?
     
  18. Jun 4, 2006 #17

    matt grime

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    As was said a long time ago pick a suitable basis of V and use the idea you just expressed in reply to AKG.
     
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