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Matrix Independence Proof

  1. Feb 4, 2010 #1
    Edit: There was a mistake in the question, see below for right question.
     
    Last edited: Feb 5, 2010
  2. jcsd
  3. Feb 4, 2010 #2
    If {C,D} was not a set of independent matrices then what sort of relation would C and D satisfy? What does taking the transpose of both sides of said relation tell you?
     
  4. Feb 4, 2010 #3
    If {C,D} were not a set of independent matrices, then I would be able to find some numbers a and b not both zero, such that aC + bD = 0.

    Transpose of this... aCT + bDT = 0?
     
  5. Feb 4, 2010 #4
    Right, now use the relations given in the question to rewrite that last equation.
     
  6. Feb 4, 2010 #5
    Okay, so that means

    aC + b(-D) = 0.

    Now that means that aC + b(-D) = aC + bD

    and can be rewritten aC -b(D) = aC + bD... which would mean that a = a, and -b = b. But the only way -b = b can be true is if b = 0. So therefore, b is zero and my initial condition is false, and they are independent!! right?
     
  7. Feb 4, 2010 #6
    oh wait, but a can still be something other than zero... my condition is that they are not BOTH zero. so my initial condition is not necessarily false =/
     
  8. Feb 4, 2010 #7
    Sort of. You can't exactly just equate coefficients if that's what you did. The last equation you wrote implies that bD = 0 by just rearranging. You can't cancel the D here either, but you can use one property given in the question which you haven't used yet. :)
     
  9. Feb 4, 2010 #8
    Wait, why can't I cancel the D? If bD = 0 and D is not zero, then isn't b zero?

    Am I supposed to use the fact that they are n x n?
     
  10. Feb 4, 2010 #9
    I just realized the question is not supposed to say that C and D are independent matrices, just that they are n x n and nonzero, if that makes a difference.
     
  11. Feb 4, 2010 #10

    Dick

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    Sure it makes a difference. You've got aC+bD=0 and aC-bD=0. To show C and D are independent, you want to show that a and b are both zero. Suppose a is nonzero?
     
  12. Feb 4, 2010 #11
    If it's still true that bD is zero, then if a is nonzero, you'd have aC is not zero.

    Then aC + bD = 0 --> aC + 0 = 0 but if aC is not zero this is obviously not true.
     
  13. Feb 4, 2010 #12

    Dick

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    Add the two equations. Who cares what bD is? If a is nonzero then you get 2aC=0. a is nonzero, C is nonzero. Possible, or not?
     
  14. Feb 4, 2010 #13
    No, not possible. Therefore a must be zero.

    But I still need to show that b is zero, right?

    Now that I know a is zero can I just plug it in to the equation aC + bD = 0?
     
  15. Feb 5, 2010 #14

    Dick

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    Sure you can. Or you could subtract the two equations and conclude 2bD=0, D being nonzero means b must be zero. Same conclusion either way.
     
  16. Feb 5, 2010 #15
    Great, so then a = b = 0, thus {C, D} is independent. And we're done!

    Thanks to both of you for your help :)
     
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