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Proving Inverse Matrices?

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm trying to show that a given matrix is the inverse of the other, by showing that multiplying them together generates the identity matrix. I can't see a way to simplify the last step and I feel like I'm missing something..? Any input on this would be helpful. Thanks!

    2. Relevant equations
    see below
    j = sqrt(-1)

    3. The attempt at a solution
    ece2200.jpg
    How do I prove that the rest of the terms are 0?
     

    Attached Files:

  2. jcsd
  3. Mar 16, 2010 #2

    LCKurtz

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    It looks to me like the last sentence, which you cut off at the comma, might give you a hint. What does it say?
     
  4. Mar 17, 2010 #3
    It's not cut off. It's as far as I have gotten with the proof, haha. There is nothing obvious that jumps out at me when I consider the case where the exponential argument isn't 0.
     
  5. Mar 17, 2010 #4

    jbunniii

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    Try using

    [tex]\sum_{m=0}^{M-1} z^m = \frac{1 - z^{M}}{1 - z}[/tex]

    which holds for any complex number [itex]z \neq 1[/itex].
     
  6. Mar 17, 2010 #5
    Thanks so much! That did the trick!

    finishedproof.jpg
     
  7. Mar 17, 2010 #6

    jbunniii

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    Looks good, nice job.

    It's also insightful to consider what is going on geometrically.

    This sum:

    [tex]\sum_{i=0}^{M-1}\frac{1}{M}e^{j\left(\frac{2\pi(\lambda-\kappa)i}{M}\right)[/tex]

    is calculating the average of [itex]M[/itex] complex numbers. Furthermore, since [itex]\lambda - \kappa[/itex] is a nonzero integer, these complex numbers are evenly spaced samples around the unit circle, so their average is zero.
     
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