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Proving Inverse Matrices?

  • Thread starter swuster
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  • #1
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Homework Statement


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!

Homework Equations


see below
j = sqrt(-1)

The Attempt at a Solution


ece2200.jpg

How do I prove that the rest of the terms are 0?
 

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Answers and Replies

  • #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?
 
  • #3
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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.
 
  • #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].
 
  • #5
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Thanks so much! That did the trick!

finishedproof.jpg
 
  • #6
jbunniii
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Thanks so much! That did the trick!

finishedproof.jpg
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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