# Proving Inverse Matrices?

1. Mar 16, 2010

### swuster

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

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

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2. Mar 16, 2010

### LCKurtz

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. Mar 17, 2010

### swuster

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. Mar 17, 2010

### jbunniii

Try using

$$\sum_{m=0}^{M-1} z^m = \frac{1 - z^{M}}{1 - z}$$

which holds for any complex number $z \neq 1$.

5. Mar 17, 2010

### swuster

Thanks so much! That did the trick!

6. Mar 17, 2010

### jbunniii

Looks good, nice job.

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

This sum:

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

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