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Vector Space Axiom: Can this be done easier?

  1. Mar 8, 2010 #1
    I think, in case it is wrong, I proved the the first vector space axiom for 3 x 3 magic squares; however, there has to be an easier way to do what I did.

    This pdf has been removed. Go to page 2 of the discussion for an updated version.

    I attached a pdf file due to I can create the document with work showing faster in Maple then using the latex feature here.
     
    Last edited: Mar 8, 2010
  2. jcsd
  3. Mar 8, 2010 #2

    vela

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    Your notation seems to be a bit messed up.

    What exactly is the problem asking you, and what can you already assume is true or known? For instance, does the problem require you to verify all the axioms explicitly? Can you assume the properties of matrix addition and scalar multiplication are known, or do you need to prove them again? Can you assume it's known that the set of 3x3 matrices (with the usual operations) form a vector space?
     
  4. Mar 8, 2010 #3
    I have to prove that all 3 x 3 magic square matrices form a vector space. I don't know of a way to generalize this without saying that the sum of all rows, columns, and 2 diagonals are equal. If I can find a way to represent it better, I would do that.
     
  5. Mar 8, 2010 #4

    vela

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    What I'm getting at is that the proof could be relatively short if you can assume you know certain facts already. For example, do you really need to show A+B=B+A explicitly or can you just say you know it's true because it holds for all matrices? Also, there is another way of proving that the set is a vector space, but it relies on already knowing the set of 3x3 matrices (with the usual operations) form a vector space.
     
  6. Mar 8, 2010 #5
    The instructions are particularly vague. It just says show it forms a vector space and find its basis.
     
  7. Mar 8, 2010 #6
    Could it be reasonable to say that the set of all 3 x 3 magic square matrices is a subset of all 3 x 3 matrices and then just prove 2 axioms addition and scalar multiplication? That would simplify the work for the axioms but that wouldn't simplify my notation.
     
  8. Mar 8, 2010 #7

    Mark44

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    That works for me.
     
  9. Mar 8, 2010 #8
    Yes - you know that the subspace inherits these properties from the larger structure. And I would imagine that the point of the exercise is to think about the specific properties of the subspace rather than prove simple vector space axioms for matrices.
     
  10. Mar 8, 2010 #9
    So I could say let V be the set of all 3 x 3 matrices and S is a subspace of V where S is the set of all 3 x 3 magic square matrices?

    Also, is there an easier way to write the set of all 3 x 3 magic square matrices in set notation w/ out the summations?
     
  11. Mar 8, 2010 #10
    Yes, you can say that S is a subset of V (and you are trying to show that it is a subspace)
     
  12. Mar 8, 2010 #11
    Does my set notation hold true then for all 3 x 3 magic squares? I wasn't entirely sure if the double summation part was correct.
     
  13. Mar 8, 2010 #12

    vela

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    You have the right idea of how to define the set of magic squares; it's just your notation was kind of weird. You could just write the condition as:

    [tex]\sum_{i=1}^3 x_{i1} = \sum_{i=1}^3 x_{i2} = \sum_{i=1}^3 x_{i3} = \sum_{i=1}^3 x_{1i} = \sum_{i=1}^3 x_{2i} = \sum_{i=1}^3 x_{3i} = \sum_{i=1}^3 x_{ii}[/tex]
     
  14. Mar 8, 2010 #13
    What about the diagonal from 31 to 22 to 13?
     
  15. Mar 8, 2010 #14

    vela

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    Oh yeah, you could add a term like

    [tex]\sum_{i=1}^3 x_{4-i,i}[/tex]

    or just write it out explicitly, [itex]x_{31}+x_{22}+x_{13}[/itex]
     
  16. Mar 8, 2010 #15
    I am not sure where to start on identifying the basis. Would I say c1....cn times each summation = 0? But then I am confused from there.
     
  17. Mar 8, 2010 #16

    Hurkyl

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    How do you normally find a basis for the solution set to a system of linear equations?
     
  18. Mar 8, 2010 #17
    I noted that above but I forget to say + each consecutive one(c1...cn multiplied by each summation and then added together = 0). If you are confused on why summation is there, that is due to not viewing the pdf. With how this appears, it becomes difficult to proceed from there though.
     
  19. Mar 8, 2010 #18
    Which allows one to find the N(A); thus, identifying the minimal span which is the basis.
     
  20. Mar 8, 2010 #19
    I just changed out the previous pdf due to an error.
     
  21. Mar 8, 2010 #20

    vela

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    You need to go back and fix your proofs. There are two major problems. First, you're not proving the right thing. Second, the way you wrote it out is an abuse of notation.
     
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