Proving Inverse Matrices: Simplifying the Last Step

[swuster]

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

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

 

[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?

 

[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.

 

[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.

 

[swuster]

Thanks so much! That did the trick!

 

[jbunniii]

Thanks so much! That did the trick!

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.