Matrix Algebra (Recurrences & Diagonalisation)

In summary: I'm having a really hard time understanding this without math. :/In summary, the problem asked for you to diagonalize an integer matrix and found the eigenvalues and eigenvectors.
  • #1
ShaunDiel
23
0
Solve this simultaneous pair of recurrences using diagonalisation
2m3iesm.jpg



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.


t6drex.jpg


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.
 
Physics news on Phys.org
  • #2
Anything ? :s
 
  • #3
ShaunDiel said:
Anything ? :s

The problem said you should use diagonalization of your matrix. First find the eigenvalues and eigenvectors.
 
  • #4
Dick said:
The problem said you should use diagonalization of your matrix. First find the eigenvalues and eigenvectors.

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
ShaunDiel said:
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

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:
  • #6
Dick said:
It does work out sort of nicely. That equation at least has integer roots. Not nasty radicals.


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
ShaunDiel said:
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?

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:
  • #8
Dick said:
You can find the eigenvectors of the integer matrix first, then divide by 143 later. 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.

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
ShaunDiel said:
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 ?

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
Dick said:
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.

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
ShaunDiel said:
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 :(

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
Dick said:
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.

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 :confused:

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
ShaunDiel said:
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 :confused:

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

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
Dick said:
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.

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

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

∴ a1 = 6
a2 = -3
 
  • #15
ShaunDiel said:
Ohh okay, so setting up the system of eq's:

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

∴ a1 = 6
a2 = -3

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
Dick said:
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).

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

aua8o4.jpg



Do I not have to do the D= P^-1 A P stuff?
 
  • #17
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
Dick said:
A(3,1)=(6/11)*(3,1). A^2(3,1)=(6/11)^2*(3,1). Use that you have eigenvectors in the expression.


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
ShaunDiel said:
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

Apply A again to both sides of those equations. Imagine what would happen if you applied A n times.
 
  • #20
Dick said:
Apply A again to both sides of those equations. Imagine what would happen if you applied A n times.

Like just multiply them?

So if I took A*[1 4] it would be [199/143 -172/143] ??
 
  • #21
ShaunDiel said:
Like just multiply them?

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

No, multiplying it out sort of misses the point. If A[1,-4]=(-5/11)[1,-4] then A(A[1,-4]))=(-5/11)*A[1,-4]=(-5/11)*(-5/11)*[1,-4]=(-5/11)^2[1,-4]. So A^2[1,-4]=(-5/11)^2*[1,-4]. What about A^3[1,-4]?
 
Last edited:
  • #22
Dick said:
No, multiplying it out sort of misses the point. If A[1,-4]=(-5/11)[1,-4] then A(A[1,-4]))=(-5/11)*A[1,-4]=(-5/11)*(-5/11)*[1,-4]=(-5/11)^[1,-4]. So A^2[1,-4]=(-5/11)^2*[1,-4]. What about A^3[1,-4]?

Would it just be A^2(A[1,-4]))=(-5/11)*A^2[1,-4]

A^3[1 -4] = (-5/11)^3[1 -4]

∴A^n[1 -4] = (-5/11)^n[1 -4]

?
 
  • #23
ShaunDiel said:
Would it just be A^2(A[1,-4]))=(-5/11)*A^2[1,-4]

A^3[1 -4] = (-5/11)^3[1 -4]

∴A^n[1 -4] = (-5/11)^n[1 -4]

?

Sure, that's it. Powers of A applied to eigenvectors are easy to compute.
 
  • #24
Dick said:
Sure, that's it. Powers of A applied to eigenvectors are easy to compute.


Sweet! So does that finish all the diagonalisation for part A?
 
  • #25
ShaunDiel said:
Sweet! So does that finish all the diagonalisation for part A?

I think so. You've found a basis of eigenvectors for A and you've expressed [c0,d0] in terms of them. Now you should be able to write down formulas for cn and dn.
 
  • #26
Dick said:
I think so. You've found a basis of eigenvectors for A and you've expressed [c0,d0] in terms of them. Now you should be able to write down formulas for cn and dn.

Okay so I have both general forms for the eigenvectors, how would I write a formula for cn & dn from those though?
 
  • #27
ShaunDiel said:
Okay so I have both general forms for the eigenvectors, how would I write a formula for cn & dn from those though?

Try it! You found [c0,d0]=6*[3,1]-3*[1,-4]. You want to find A^n of that. You know how to find A^n of the eigenvectors, right?
 
  • #28
Dick said:
Try it! You found [c0,d0]=6*[3,1]-3*[1,-4]. You want to find A^n of that. You know how to find A^n of the eigenvectors, right?


Do I just do it the same way as before? Applying A to both sides?

So: A[c0,d0] = =6*A[3 1] -3 *A[1 -4]
 
  • #29
ShaunDiel said:
Do I just do it the same way as before? Applying A to both sides?

So: A[c0,d0] = =6*A[3 1] -3 *A[1 -4]

Sure. But why not apply A^n to both sides? The left side becomes [cn,dn] doesn't it?
 
  • #30
Dick said:
Sure. But why not apply A^n to both sides? The left side becomes [cn,dn] doesn't it?


Oh damn, so it does.

So I'll have :


[cn dn] = 6*A^n[3 1] -3 *A^n[1 -4]

as the expression for cn & dn? or should they be seperate?

I'm guessing separate since it asks for a ratio in part b)
 
  • #31
ShaunDiel said:
Oh damn, so it does.

So I'll have :


[cn dn] = 6*A^n[3 1] -3 *A^n[1 -4]

as the expression for cn & dn? or should they be seperate?

I'm guessing separate since it asks for a ratio in part b)

Keep going. You know expressions for A^n[3,1] and A^n[1,-4]. Then equate the components to find cn and dn separately.
 
  • #32
UGH, I'm pretty sure I messed up way back around the start.

http://www.wolframalpha.com/input/?i=[[1%2F143*67%2C1%2F143*33]%2C[1%2F143*44%2C-1%2F143*54]]

My eigenvalues are right, but 1 vectors wrong.
 
  • #33
Dick said:
Keep going. You know expressions for A^n[3,1] and A^n[1,-4]. Then equate the components to find cn and dn separately.


Okayy, So I'lll have:

[Cn, dn] = 6 (6/11)^n *[3 1] - 3 (-5/11)^n * [1 -4]

What do you mean by equate components?
 
  • #34
You have a vector equal to a vector. They are equal if each element of one vector is equal the the corresponding element of the other vector. So cn=?? You know, I really don't think are are so bad at this, you found the eigenvectors easily and found the eigenvalues without much trouble. I don't know why you are checking back for approval of basic things so often. The next question that comes up, why don't you try and imagine what I would answer and then jump to the next step?
 

1. What is matrix algebra?

Matrix algebra is a branch of mathematics that deals with the study of matrices, which are arrays of numbers or symbols arranged in rows and columns. It involves operations such as addition, subtraction, multiplication, and division of matrices, as well as solving systems of linear equations using matrices.

2. What is the purpose of using matrix algebra?

The purpose of using matrix algebra is to simplify and solve complex mathematical problems involving large sets of data. It is commonly used in fields such as engineering, physics, economics, and computer science to model and analyze real-world systems.

3. What are recurrences in matrix algebra?

Recurrences in matrix algebra refer to a sequence of equations that define the elements of a matrix based on the values of its previous elements. This is often used in iterative algorithms to solve problems involving matrices.

4. What is diagonalisation in matrix algebra?

Diagonalisation in matrix algebra refers to the process of finding a diagonal matrix that is similar to a given matrix. This is useful in simplifying calculations and solving systems of linear equations, as well as in identifying patterns and relationships within the data represented by the matrix.

5. How is matrix algebra used in data analysis?

Matrix algebra is an essential tool in data analysis as it allows for the manipulation and transformation of large datasets. It is used to perform operations such as data scaling, normalization, and dimensionality reduction, as well as in the creation of statistical models and machine learning algorithms.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
463
  • Calculus and Beyond Homework Help
Replies
4
Views
640
  • Calculus and Beyond Homework Help
Replies
25
Views
2K
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top