# Matrix Algebra (Recurrences & Diagonalisation)

1. Apr 25, 2012

### ShaunDiel

Solve this simultaneous pair of recurrences using diagonalisation

Not sure what would be related equations to this.

Originally I had no idea how to do this, I set up the first matrix, like this.

Then, from there, I know that I have to let:

x_k= [[c_k][d_k]]. Then x_k = A^k x_0,so you want powers of A. If you find its eigenvectors, then A = P * [[lambda_1, 0 ] [ 0, lambda_2]] * P^-1, so A^k is P*[[lambda_1^k, 0 ] [ 0, lambda_2^k]]*P^-1.

I just don't understand the syntax in this and how to put it to paper

Please give me a push in the right direction, I've been stuck on this problem all day.

2. Apr 25, 2012

### ShaunDiel

Anything ? :s

3. Apr 25, 2012

### Dick

The problem said you should use diagonalization of your matrix. First find the eigenvalues and eigenvectors.

4. Apr 25, 2012

### ShaunDiel

Is A the correct matrix to diagonalise though? I did it out right now and usually on test problems our prof makes everything work out fairly nicely but this worked out to be:

λ^2 - 13λ -5070

Divided by 143 in front

5. Apr 25, 2012

### Dick

It does work out sort of nicely. That equation at least has integer roots. Not nasty radicals. If this is a test question, I probably would have tried to find something nicer though.

Last edited: Apr 25, 2012
6. Apr 25, 2012

### ShaunDiel

Oh damn, it factors too,

So I have my two Eigenvalues:

λ1= 78
λ2= -65

Should these be divided by 143? Or Can I leave that alone to solve for my eigenvectors?

7. Apr 25, 2012

### Dick

Find the eigenvectors of the integer matrix first. Do you know what you are going to do with the eigenvalues and eigenvectors once you find them? It doesn't hurt to look ahead.

Last edited: Apr 25, 2012
8. Apr 25, 2012

### ShaunDiel

Well, I'm solving for my Eigenvectors now, All 4 numbers in both matrices are multiples of 11, after that would I do the:

D= P^-1 A P ?

9. Apr 25, 2012

### Dick

That was the point to the "looking ahead" idea. If your initial condition are an eigenvector, then you know what happens. If it a mix of the largest eigenvalue and the smaller eigenvalue then you should also know what will happen.

10. Apr 25, 2012

### ShaunDiel

This is where I start to get lost,

I 'solved' for my eigenvectors by plugging in λ1 & λ2, Which resulted in 2, 2x2 matrices, simple row ops removed a row yielding :

1) 1/143 [1 3]
2) 1/143 [-4 1]

So [x,y] = (3,1)t & (1,-4)t

Now I'm lost :(

11. Apr 26, 2012

### Dick

Ok, so you've got the two eigenvectors. v1=(3,1) and v2=(1,-4) and they correspond to the eigenvalues λ1=78/143=6/11 and λ2=-65/143=-5/11. You initial state is (15,18). Can you express that as a linear combination of your eigenvectors? Now it should be easy to say what cn and dn are.

12. Apr 26, 2012

### ShaunDiel

I have no idea how to do this at all, I've been watching youtube videos for the past hour trying to figure it out and there's nothing in my textbook regarding recurrences

EDIT: Does it have anything to do with the initial conditions being multiples of 3 of my eigenvalues or is that just a coincidence?

13. Apr 26, 2012

### Dick

You just want to solve a1*(3,1)+a2*(1,-4)=(15,18) for a1 and a2. If you express it in components it's two equation in the two unknowns a1 and a2.

14. Apr 26, 2012

### ShaunDiel

Ohh okay, so setting up the system of eq's:

3a1 + a2 = 15
a1 - 4a2 = 18

∴ a1 = 6
a2 = -3

15. Apr 26, 2012

### Dick

Fine. So now you've expressed (c0,d0)=6*(3,1)-3*(1,-4). Since (3,1) and (1,-4) are eigenvectors of A, it should be easy to write down an expression for (cn,dn)=A^n(c0,d0).

16. Apr 26, 2012

### ShaunDiel

I'm sorry, I feel like I'm useless at this..

Do I not have to do the D= P^-1 A P stuff?

17. Apr 26, 2012

### Dick

A(3,1)=(6/11)*(3,1). A^2(3,1)=(6/11)^2*(3,1). Use that you have eigenvectors in the expression.

18. Apr 26, 2012

### ShaunDiel

I just really don't understand what's going on.. I have

A[1,-4] = (-5/11)*[1 -4]
A[3 1] = (6/11) & [3 1]

But what am I trying to do? Get what in terms of what? :s Thanks for being so patient with me I feel like an idiot

19. Apr 26, 2012

### Dick

Apply A again to both sides of those equations. Imagine what would happen if you applied A n times.

20. Apr 26, 2012

### ShaunDiel

Like just multiply them?

So if I took A*[1 4] it would be [199/143 -172/143] ??