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Linear Algebra question

  1. May 7, 2006 #1
    Let the set S be a set of linearly independent vectors in V, and let T be a linear transformation from V into V. Prove that the set
    [tex]
    {T(v_1), T(v_2),...,T(v_n)}
    [/tex]
    is linearly independent.

    We know that any linear combination of the vectors in S, set equal to zero, has only the trivial solution. We use the linear combination of all vectors in S and then take the linear transformation T of that vector. Then we can write this linear combination as a linear combination
    [tex]
    c_1T(v_1) + c_2T(v_2)+ ... + c_nT(v_n)
    [/tex]

    We need to show that if we make the above linear combination a homogenous equation, then the only solution is the trivial solution. But that's where I'm stuck. How do we know that S is not in the kernal of T? If it is, then where do we go? Any idea's on where to go are appericated.

    TNA
     
  2. jcsd
  3. May 7, 2006 #2

    matt grime

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    You don't know that some element in the span of S is not in the kernel. Are you sure you stated the whole problem correctly?
     
  4. May 7, 2006 #3
    I stated the problem verbatim from the text...

    I'm thinking the key is that it's from V into V. If a set of linearly independent vectors are a subset of V, then the span(S) is a subspace of V. Do we also know then that T of a linear combination of the vectors S is a subspace of V? Clearly T contains the zero vector, and we know that since T maps from V to V that T is close under addition and scalar multiplication. So yeah, I guess T is also a subspace of V.

    So I know that span(S) is a subspace of V if not all of V, and I know that span(T(v)) is a subspace of V. I know that S is a linearly independent set. But again, the whole Kernal issue makes we wonder how I'm supposed to know that T is linearly independent....

    I'm so lost on this problem...I'm just afraid this problem might be on the final, Monday morning, so I really want to figure it out.

    Again, thanks in advance.
     
  5. May 7, 2006 #4

    LeonhardEuler

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    Something is definitely wrong about this problem. What if T is the zero transformation? Then T(v1)=t(v2)=...=T(vn)=0, so the T(vi) are obviously not linearly independant. Are you sure the problem didn't say something like "let T be an invertible linear transformation" or something like that?
     
  6. May 7, 2006 #5
    I think I've got it....

    The only way a set S of vectors is linearly dependent is if there some vector in S that can be written as a linear combination of the other vectors in S.

    Assume T is linearly dependent for some vector in S. Then WLOG there exist some T(v) such that
    [tex]
    T(v_i) =c_1T(v_1)+ ...+c_{i-1}T(v_{i-1})+...+c_nT(v_n)
    [/tex]

    Which can be rewritten as
    [tex]
    c_1T(v_1)+ ...+c_{i-1}T(v_{i-1}) - T(v_i)+...+c_nT(v_n)=0
    [/tex]

    Which then can be written as the linearly transformation of the set of vectors S. This implies that the linear combination of vectors in S has a nontrivial solution becasue the coefficent of v_i is not zero, but instead -1. This is absurd because S is linearly independent. Therefore T of the linear combination of vectors in S is linearly independent.

    Would that be good enough?

    EDIT:
    Oh, and I guess I would need to add that since T is from V into V, the zero vector is a fixed point for T going from V into V. Which is exactly why I can say that the linear combination that maps to the 0 vector is exactly equal to the zero vector.
     
    Last edited: May 7, 2006
  7. May 7, 2006 #6
    I'm totally sure...the problem is stated exactly as it is written in the text.

    The text is:
    Elementary Linear Algebra 5th edition, Larson/Edwards/Falvo.
    Page 364 problem # 56. The chapter is 6.1.....

    In any case, I think I might have it....let me know if you think my proof works. Thanks
     
  8. May 7, 2006 #7

    matt grime

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    I didn't appreciate the word 'into' was there. That means it is injective, so the kernel is trivial. (Into is strictly weaker than invertible since V might be infinite dimensional.)

    Your proof doesn't appear to actually use that fact, though there are some other problems than that ("Assume T is linearly dependent for some vector in S" for instance doesn't make sense), so I don't think it is correct.
     
    Last edited: May 7, 2006
  9. May 7, 2006 #8

    LeonhardEuler

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    I knew the word "onto" was synonymous with injective, but is the same true of "into"? I've never seen that word used that way, but maybe I'm just not familiar with it.
     
  10. May 7, 2006 #9
    I added the fact that T(0)=0 is a fixed point for T:V->V. Meaning that whatever the vector that goes into T is, if T of that vector is the 0 vector, then that vector is also the zero vector. Does that resolve the other problems? Should I also need to prove that 0 is a fixed point for T since it's not explictly stated as a given in the problem?
     
  11. May 7, 2006 #10

    LeonhardEuler

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    The beginning of the problem says "Let S be a set of linearly independant vectors in V". If one of them was 0, then the set would be linearly dependant, so you know that none of them are zero, so 0 poses no problem.
     
  12. May 7, 2006 #11

    matt grime

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    onto is not at all synonymous with injective: it is surjective.
     
  13. May 7, 2006 #12

    LeonhardEuler

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    Oh, right.
     
  14. May 7, 2006 #13
    I realize that, but the problem I was having was in knowing wheather or not any of the vectors were in the kernal of T. Since T(v)=0 is a fixed point I then know that v=0. That's what a fixed point is. In other words, the Kernal of T is just the zero vector. From that knowledge I know that if T(linear combination)=0, then linear combination =0. Since the linear combination is linearly independent, I can then say that the linear combination of T, as stated above, is going to be linearly indepent by using properties of a linear combinations. Would you agree, or no?:smile:
     
  15. May 7, 2006 #14

    matt grime

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    that doesn't follow at all. Tx=0 for all x has 0 as a fixed point but fails your assertion. ('that goes into T is' doesn't make sense)

    no


    it is given: T is linear.
     
  16. May 7, 2006 #15

    matt grime

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    no this does not follow at all. fixed points have nothing to do with anything.

    you have just asserted that every linear map has trivial kernel.

    no
     
  17. May 7, 2006 #16

    matt grime

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    The use of the word 'into' to mean injective is the only way that this question can be 'true'. It is not often done (I can think of no example where I definitely know someone uses into to mean injective, but I have a feeling that people do use it, possibly in mistranslation).

    For the OP.

    Let T be the zero map, let S be a basis of V, then T(S) is not a linearly independent set. You *must* use the injectivity to conclude that the only vector mapped to zero is the zero vector, and saying 0 is a fixed point doesn't do anything.
     
  18. May 7, 2006 #17
    A fixed point is such that T(v)=v.

    Proof that 0 is a fixed point of any linear transformation T:V->V.

    Because T(0)=0 for any linear transformation T, 0 is a fixed point of T.

    Therefore
    [tex]
    0=c_1T(v_1)+...+c_nT(v_n)
    [/tex]

    implies that the linear combination of S is the zero vector. :confused:

    I really don't like to disagree with a math guru, but I don't see the problem here. What more should I need to make the proof complete?

    Geekster (has been humbled too many times to want to disagree with math gurus)
     
  19. May 7, 2006 #18

    matt grime

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    So you are happy that you've just proved every linear map is injective? Even the zero map?

    Let T be the zero map. T(sum c_iv_i) is always zero irrespective of what sum c_iv_i is, so you can clearly see that what you've written doesn't imply that sum c_iv_i is zero.
     
  20. May 7, 2006 #19

    LeonhardEuler

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    I think you're phrasing this wierd, but you have the idea. If I understand you, you're saying that, suppose you have a linear comination of T(v1), T(v2)..., T(vn) that was equal to zero. Then use the linearity properties to convert this to T(c1v1+c2v2+...c3v3)=0. Then use the fact that T(v)=0 implies v=0 to get c1v1+c2v2+...c3v3=0. Then I loose you. You say "Since the linear combination is linearly independent, I can then say that the linear combination of T, as stated above, is going to be linearly indepent by using properties of a linear combinations." I don't know what you mean, but anyway you're 95% finished the problem at this point. Think about what the equation c1v1+c2v2+...c3v3=0 says about the ci.
     
  21. May 7, 2006 #20
    I see....

    I'm thinking perhaps we've found an error in the text then. Everything seems to lead upto the idea that I can use a fixed point to prove that T is linearly independent since the concept of an injective function has not yet been mentioned. How would we know that T is in fact injective? After all, T could be the zero map like M.G. mentioned....

    And more importantly....

    IF this problem is on my finial, and the teacher expects me to use the fixed point to prove that T is LI, when this is not true in general, what should I do?

    And by the way, in the context of the textbook I am using 'into' just means that the function maps to V. In other words, T maps V into V, just means that T is a subspace of V, nothing more.

    I think I might be screwed for this test.....the test is comprehensive and there are just so many possibilities for questions on this test.....If I get a proof question like this on the test, I'll spend all my time stuck right there...:cry:

    I dunno....honestly I understand what you're saying Matt G. and at this point I don't think enough information is given in the problem to do the proof. If anyone has the instructors solution manual (teachers) perhaps they could let us know what the authors think the solution is.
     
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