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Homework Help: Spanning proof

  1. Jan 20, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm having a lot of trouble with this question. I don't even know where to begin.
    span{X1, X2, ... ,Xq}. a matrix A is m x n, and it is a zero matrix. Prove that AXp is NOT zero for some p.


    2. Relevant equations



    3. The attempt at a solution
    I'm guessing I need to show that Xi is never zero or something, because that way AXi will also not be zero. Since the span there is equal to aX1 + bX2 + ... + qXq, it doesn't make sense that Xi would be zero ever.

    I'm sure that's wrong, because there are cases where two non zero matrices multiplied give a zero matrix. Also, I think my proof should be more mathematical. Can anyone steer me in the right direction?
     
    Last edited: Jan 21, 2010
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  3. Jan 20, 2010 #2

    Dick

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    What have {X1, X2, ... Xq} got to do with the matrix A?? I think you've left a lot of the question out. If the span of the X's is in the null space of A that's certainly not true.
     
  4. Jan 20, 2010 #3
    Reading the question again, the only thing I left out was that the R should be to the power of n.

    The only link I can see between A and span{X1, X2, ... , Xq} is that any of those Xs multiplied by A cannot give zero.

    Also the question doesn't mention the null space of A.
     
  5. Jan 20, 2010 #4

    Dick

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    That helps a lot. So the X's span all of R^n. Define f(x)=Ax. f(x) maps R^n->R^m. If A is nonzero, can the range of f(x) be the zero vector in R^m?
     
  6. Jan 20, 2010 #5
    Okay, my first problem is: what's the difference between R^n and R^m? R^n is all the columns? I'm pretty new to all of this.

    I'm guessing the range of f(x) can't be the zero vector?
     
  7. Jan 20, 2010 #6

    Dick

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    I'm just not using the funky font to write R. R^n is the same space you referred to in your original post. It's the space of all column vectors with n components. Similar for R^m. So you can see how Ax maps R^n to R^m, right?
     
  8. Jan 20, 2010 #7
    oh, I think I might see it now. if A is non zero then there can't be any zero vectors in R^m OR R^n, right? So f(x) = Ax means that x can't be zero. is that right?

    if so where are you getting f(x) = Ax though? is that a given fact?
     
  9. Jan 20, 2010 #8

    Dick

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    No, you aren't getting it. Of course there's is a zero vector in R^n and R^m. It's the column filled with zeros. I'm just writing f(x)=Ax to emphasize Ax is a function of x. If x is a column vector in R^n then Ax is a column vector R^m. Are you following that?
     
  10. Jan 20, 2010 #9
    okay yes, that makes sense. because when you multiply a column vector with n components by A, you get a column vector with m components. correct?
     
  11. Jan 20, 2010 #10

    Dick

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    Ok, now can you PROVE that if A is nonzero then there is SOME vector in R^n such that Ax is nonzero? That's half the proof. Hint: if A is nonzero then some column must contain a nonzero entry. Suppose it's column number k. Can you think of a vector in R^n such that Ax is nonzero?
     
  12. Jan 20, 2010 #11
    Wouldn't any vector except the zero vector in R^n give an Ax that is nonzero?
     
  13. Jan 20, 2010 #12

    Dick

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    No! You aren't really visualizing this. Suppose A is the 2x2 matrix [[1,0],[0,0]]. I.e. first row 1 0 second row 0 0. Ok? A is nonzero, right? Take x to be the 2x1 column vector [0,1]. I.e 0 on the top and 1 on the bottom. x is nonzero, also right? What is Ax?
     
  14. Jan 20, 2010 #13
    Ax is [0,0].

    but obviously there are some vectors that do not give [0,0], like if x is [2,3]... I don't know how to actually prove that though!
     
    Last edited: Jan 20, 2010
  15. Jan 20, 2010 #14

    Dick

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    That was my hint. If column number k of A contains a nonzero entry and you define a column vector x to be all zeros except for a 1 in row k, can Ax be zero? Are you still visualizing matrix multiplication? What does Ax have to do with the columns in A?
     
  16. Jan 20, 2010 #15
    No, Ax cannot be zero, you will always get a one in row k of Ax.

    And... if there are 2 columns in A, then Ax has that many rows?
     
  17. Jan 20, 2010 #16

    Dick

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    Hmm. I'm going to hope you are seeing this. If x is as I described it before, would you agree Ax IS column number k of A??
     
  18. Jan 20, 2010 #17
    wow, I definitely didn't see that before! but yes, I would agree with that.
     
  19. Jan 20, 2010 #18

    Dick

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    Ok, we're getting there! So if A is nonzero, then it must have a nonzero column. If A has a nonzero column you can pick a vector in R^n such Ax is that column, so Ax is nonzero. Now your question says span{X1,...Xq} is R^n. So x=a1*X1+...+aq*Xq for some choice of numbers a1...aq. Can you see where this is going?
     
  20. Jan 20, 2010 #19
    Nope, I'm having trouble making the connection now as to why x is equal to span{X1... Xq}.

    I'm guessing but that one entry in A which is non zero would be multiplied by x?
     
  21. Jan 20, 2010 #20

    Dick

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    No, no, no again. x isn't EQUAL to span{X1...Xq}. The span is a SET of vectors. x is a single vector. x is IN the span. Please think hard about this, ok? Remember they told you span{X1...Xq} is EQUAL to R^n. Is x in R^n?
     
  22. Jan 20, 2010 #21
    I thought the definition of span{X1 ... Xq} was a1x1 + ... q1Xq though =/ My prof didn't really explain what span was, so maybe I'm unclear on that.

    x is in R^n... so therefore x is in span{X1 ... Xq}.
     
  23. Jan 20, 2010 #22

    Dick

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    Try to get clear on that. span{X1...Xq} is the set of ALL vectors that can be written as a1*X1+...aq*Xq where the a's range over ALL their possible values. So yes, if x is in R^n and R^n is the span{X1...Xq} then x can be written in the form x=a1*X1+...+aq*Xn for SOME choice of a1...aq. Is this making sense?
     
  24. Jan 20, 2010 #23
    oh okay, so it doesn't mean x is equal to the span of x, but rather equal to one particular linear combination of the vectors that span R^n, right?
     
  25. Jan 20, 2010 #24

    Dick

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    Right, right, right! I think you are getting it. And X1...Xq span R^n. I wouldn't say 'one particular', but you can say 'at least one'. I'd be ok with that.
     
  26. Jan 20, 2010 #25
    okay, that makes sense to me now! so going back to the proof, that vector x is in the span, so Ax = Aa1X1 + ... + AqXq?
     
    Last edited: Jan 20, 2010
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