# Matrix General Expression

1. Oct 3, 2006

### lapo3399

Hi,

I have a problem which involves determining a general expression in terms of k and n for the exponentiation of a matrix of the form:

The general expression I determined was accurate through a seperate process (which does not concern this post) was:

I am supposed to prove this expression is correct symbolically (ie I cannot use any examples with inputted k values) and I believe that the following should suffice:

Does this seem correct? Also, is there a way that I can prove this without inputting values of n (ie entirely symbolical)?

Any help is greatly appreciated.

Matt

2. Oct 3, 2006

### StatusX

Try induction.

3. Oct 3, 2006

### lapo3399

Unfortunately, I am in Grade 12 Discrete and we haven't done mathematical induction yet. I understand the basics of induction, but I have no idea how it should work with matrices and two variables. If anyone could help me with this, I would be grateful.

4. Oct 3, 2006

### shmoe

You would only being doing induction on the one variable "n". The set-up the the same as usual,

base case: prove X^1 has that form (done)
induction step: assuming your formula for X^n is correct, prove it is correct for X^(n+1). Just use X^(n+1)=X*(X^n) and use your formula for X^n.

5. Oct 3, 2006

### lapo3399

I believe this is what I am supposed to do:
Proving the expression works for X^1, assuming that it is correct for X^n and then proving it is correct for X^(n+1):

However, there seems to be a problem.... the last 2 is to the exponent n+1, when it should be to the power of n, shouldn't it?

6. Oct 3, 2006

### shmoe

There's some funny business going on there. To the right of your "X^(n+1)=..." it looks like you have X times your expression for X^(n+1), not X^n, since it has k^(n+1)'s all over and a 2^n in front.

Then you had some problems when multiplying by this X two lines later. This may be some kind of "off by one" transcription error, but check it over carefully.

7. Oct 3, 2006

### lapo3399

I see the problem... I did it almost correctly on paper, and the only thing wrong with the equations I made was that I accidentally wrote in the n+1 (which I didnt do on paper :uhh: ) In any case, the real problem was that I DID substitute n+1 in for n in the exponent of the 2 when multiplying, so that error carried down.
Thanks for all the help!
p.s. Is there a way to prove it works for all negative integers too? or does this prove that?

8. Oct 3, 2006

### shmoe

This only proves it for positive n. For negative integers, just invert what you have. This only makes sense when X is invertable of course.

9. Oct 3, 2006

### lapo3399

So do it with n-1 instead of n+1?

10. Oct 3, 2006

### shmoe

I mean if n is positive, to find X^(-n) just take your formula for X^n and find the matrix inverse.

11. Oct 3, 2006

### lapo3399

I'm sorry but I don't understand - what do I do once I find the inverse? Find if it's true for n-1? And if so, assuming what?

12. Oct 3, 2006

### shmoe

Once you find the inverse you are done. Invert the formula for X^n and you get a formula for X^(-n). Since you've proven your formula for X^n was true for all positive n=1,2,3,4... this gives you a formula for all negative exponents n=-1,-2,-3... in one fell swoop.

Ok, you're not quite done. It would be nice to show your formula for X^n works when n=0 as well, but that's simple enough to do. Now you will have a formula for X^n that's valid for all integers n.

13. Oct 3, 2006

### lapo3399

Easy enough.... Thank you shmoe, you have been most helpful :D!

14. Oct 3, 2006

### shmoe

Happy to help.

By the way, good show on putting down all your work in a legible way . Since you are relatively new, you may not be aware that you can use latex here, as in

$$X=\left[\begin{array}{cc}k+1 & k-1\\k-1 & k+1\end{array}\right]$$

(click on the image to see what you need to type to produce it).

What you did was perfectly fine though! I only point it out as another option.

15. Oct 3, 2006

### lapo3399

I did, but thanks for reminding me. This is part of a report for school, so I had the equations (well, most of them) previously made in that format anyway. But thanks for letting me know

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?