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Span Proof help !

  1. Oct 14, 2006 #1
    Span Proof help plz!

    Hi,

    I'm working through a textbook on Linear Algebra and just need some info on some proofs.

    The text gives the Theorem but only provides proof on only one of the parts and I'm really curious as to how to prove b) and also c) so that I can understand the Theorem a bit better. It says the proofs are easy but I've just started learning on this topic so I'm not too clued up :redface: I've looked around on the internet and other books for proofs on this but can't find anything :frown:

    Can anyone give me a slight briefing on the proofs?

    Many many thanks!
     
    Last edited: Oct 15, 2006
  2. jcsd
  3. Oct 14, 2006 #2

    quasar987

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    for b)

    denote [itex]A=\left\{v_1,v_2,...,v_n \right\}[/itex], [itex]B=\left\{v_1,v_2,...,v_n,...,v_{n+m} \right\}[/itex]

    Then u in sp(A) is of the form

    [tex]u=\sum_{i=1}^n c_iv_i = \sum_i^{n+m}c_iv_i[/tex]

    where [itex]c_i =0 \ \forall i\geq n+1[/itex]. Nevertheless, this shows that u is in sp(B) also.
     
    Last edited: Oct 14, 2006
  4. Oct 14, 2006 #3

    quasar987

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    for c)

    subspace have the property of closure. I.e if u,v are in A, then au+bv is in A also. I.e any linear combination of vectors of A is in A also. But that is all sp(A) is: the set of all linear combinations A. So since the element of sp(A) are all in A, [itex]sp(A)\subset A[/itex].

    And according to the first part of the thm, [itex]A\subset sp(A)[/itex].

    These two relations btw set can only happen if they are equal: A=sp(A).
     
  5. Oct 14, 2006 #4
    Thanks very much for the proofs!! Very clear to understand :) I'm still trying to understand the one for b) but I'm getting there :tongue2: c) is very clear to me now and has confirmed the first half of my own attempted proof so it does reassure me that I am taking some of this in :rolleyes:

    I came across this interesting one and was wondering if you had the time to help me out on this one too?

    http://i9.tinypic.com/4dfhoie.jpg

    Your help very much appreciated :smile:
     
  6. Oct 14, 2006 #5

    quasar987

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    In this one, V is a vector space and U,W are subspaces. The author asks "Is there a largest subspace of V contained in both U and W?".

    If you just look and U and W as sets, then [itex]U\cap W[/itex] is the largest set contained both in U and W (of course it is; [itex]U\cap W[/itex] is defined as the set of all the elements common to both U and W). Now we ask: is [itex]U\cap W[/itex] a subspace? If it is, then it is the largest subspace of V contained in both U and W.

    To show that it is a subspace, consider u,v in [itex]U\cap W[/itex] and a,b two scalars. Then u and v are in both U and W (by definition of intersection). Both both U and W are subspaces, so they have the property of closure, meaning that au+bv is in both U and W. So au+bv is in [itex]U\cap W[/itex] (by definition of intersection).
     
  7. Oct 15, 2006 #6

    JasonRox

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    Whoa! I'm so confused.

    What is b) and c)?

    quasar987 is one hell of a mind reader!
     
  8. Oct 15, 2006 #7

    quasar987

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    lol, there was a file attached in the OP.
     
  9. Oct 15, 2006 #8
    LOL!! Sorry I took the link out because the website only allows images to uploaded for a certain period of time and since my question has been answered, I took it out :redface: sorry!

    Okay I have just one more question, if you have the time to help me out. Thanks very much for being so patient with me lol! As you can see I am a beginner and I'm finding linear algebra very tedious :frown:

    It has stated as an example but the author has not gone into an explanation so I'm kind of confused and curious at the same time. I don't really know how to start proving or showing how such and such is a subspace when it concerns matrices :confused:

    EDITED: Thanks again!
     
    Last edited: Oct 15, 2006
  10. Oct 15, 2006 #9

    quasar987

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    An easy test to check if a subset of a vector space is a subspace is to check whether the null element is in it. Suppose V is a vector space and A is a subset of V, but A does not contain 0. Then A is not a subspace because subspaces have the property of closure, meaning that for any v in A and constant c, cv is in A. In particular, for c=0, the closure property says that 0v=0 is in A. So you see that if 0 is not in A, A cannot have the closure property and thus cannot be a subspace.

    Now consider [itex]P_2'[/itex]={[itex] p\in P_2:a\neq 0\left\}[/itex]. Since a[itex]\neq[/itex]0, the null polynomial (0t+0), is not in [itex]P_2'[/itex].

    And so, with that in mind, if we consider [itex]M_{2\times 2}(P_2')[/itex], the set of 2x2 matrices whose entries are elements of [itex]P_2'[/itex], then the null matrix [tex]\left( \begin {array} {cc} 0 & 0 \\ 0 & 0 \end{array}\right)[/tex] is not in [itex]M_{2\times 2}(P_2')[/itex]. So [itex]M_{2\times 2}(P_2')[/itex] is not a subspace of [itex]M_{2\times 2}(P)[/itex].
     
    Last edited: Oct 15, 2006
  11. Oct 15, 2006 #10
    Hmm okay I think I'll have to think over that carefully...just out of curiosity, what would the 2x2 matrix [itex]M_{2\times 2}(P_2')[/itex], look like if I had to write it out? Like what would the elements be called compared to [itex]M_{2\times 2}(P)[/itex], matrix? :confused:
     
  12. Oct 15, 2006 #11

    quasar987

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    Well if I understood the book correctly, P_2 is the set of all first degree polynomials (why didn't he call it P_1?), so the elements of [itex]M_{2\times 2}(P_2)[/itex] are of the form

    [tex]\left( \begin {array} {cc} a_1t+b_1 & a_2t+b_2 \\ a_3t+b_3 & a_4t+b_4 \end{array}\right)[/tex]
     
  13. Oct 15, 2006 #12
    hmm good question! but thanks! i think i'll stop with the questions for now and have a good read over the stuff again. Thanks again! :biggrin:
     
  14. Oct 15, 2006 #13

    JasonRox

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    I would just check for closure of scalar multiplication. I never bother looking for the zero vector.
     
  15. Oct 15, 2006 #14

    JasonRox

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    I believe most books use P_2 to be the set of all first degree polynomials. I'm not 100% sure, but for some reason I doubt it. Maybe the author was looking for an easy way to denote the vector space while making the dimension of the space obvious.

    I remember once seeing something of the sort like...

    The dimension of P_n is n+1.

    Which is precisely what you think it should be.
     
  16. Oct 15, 2006 #15
    yeah scalar multiplication is also a good way of checking...
     
    Last edited: Oct 15, 2006
  17. Oct 15, 2006 #16

    quasar987

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    A scalar is just a number. The test for closure under scalar multiplication would be checking if for any real number c,

    [tex]c\left( \begin {array} {cc} a_1t+b_1 & a_2t+b_2 \\ a_3t+b_3 & a_4t+b_4 \end{array}\right)\in M_{2\times 2}(P_2')[/tex]

    But obviously, for c=0, the resulting matrix (the null matrix) is not in P_2', because the null polynomial is not in P_2'. However for any other scalar c, the abover matrix is in [itex]M_{2\times 2}(P_2')[/tex]..
     
    Last edited: Oct 15, 2006
  18. Oct 15, 2006 #17
    lol yeah i had to re-read JasonRox's post and realised it was scalar multiplication :redface:

    There's an example here in the book where the author has taken two matrices, added them together and then multiplied them by a scalar, hence concluding that the set is a subspace of the matrix 2x2...is that necessary to see if it satisfies the addition property as well? :confused:
     
  19. Oct 15, 2006 #18

    JasonRox

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    It's not in P_2?

    I always thought 0+0t was a polynomial of degree 1. I'm sure it is.
     
  20. Oct 15, 2006 #19

    JasonRox

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    Yes, you must check both closure by scalar multiplication and addition.

    This should be a theorem in your text.
     
  21. Oct 15, 2006 #20
    Ohh yeah I found it :biggrin:
     
    Last edited: Oct 16, 2006
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