# Exam question help

1. Aug 8, 2008

### franky2727

two part question the first part i understand the second part i dont

first part is find the eigenvectors and values of the matrix (4,-1 then below-4,4)

giving me lamda=2 (1,2)T and lamda=6 (1,-2)T

after this we have part 2 that i dont really understand how to do, ive done it before but have since forgot how to and cant find the corresponding notes

B Use this result to find the sollution to the two dimensional linear map

Xn+1= 4Xn-Yn

Yn+1=-4Xn+4Yn

with X0=1 and Y0=1

thanks in advance for help, i also have a similar question but without the limits which to be honest i have never fully understood so i'll try to wrap my head arround this one then post that one after

2. Aug 8, 2008

### konthelion

Since say, let A is a square matrix i.e. 2x2 matrix, then let

$$x_{n+1} = 4x_{n} - y_{n}$$ (1)
$$y_{n+1} = -4x_{n}+4y_{n}$$ (2)

then this is a recursive function, notice that the coefficients are equal to your original matrix.
================================================
Hint:
Let A be an nxn matrix. Let $$r_{k+1} = Ar_{k} , k=(0,1,2....)$$

then to find a solution then take an eigenvector say $$r_{0}$$ and its eigenvalue $$\lambda$$ and let

$$r_{k}= \lambda ^{k}x_{0}$$

Another hint: Let $$r_{0} = c_{1}v{1}+c_{2}v_{2}= [v_{1} v_{2}] \begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix}$$
What are the constants c? What are the vectors v1, v2?

Last edited: Aug 8, 2008
3. Aug 8, 2008

### franky2727

sorry still totaly lost, could you go through it step by step so i can see whats going on?

4. Aug 8, 2008

### konthelion

Couple of typos on my initial post. Reread it along with this post.

$$A= \begin{bmatrix} 4 & -1 \\4 & 4 \end{bmatrix}$$

and that you are given $$r_{0} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

let $$v_{1}$$ be your first eigenvector, and $$v_{2}$$ be your second eigenvector.

Notice that $$Av_{1}=\lambda_{1}v_{1}$$ and $$Av_{2} = \lambda_{2}v_{2}$$

Then

$$\boxed{r_{k}=c_{1} \lambda_{1}^k v_{1} + c_{2} \lambda_{2}^k v_{2}}$$

Remember, notice that $$Ar_{k} = A(\lambda)^kr_{0} = \lambda^k(Ar_{0}) =\lambda^k (\lambda r_{0}) = \lambda^{k+1}r_{0} = r_{k+1}$$
Hope that helps
===============================
Your first job is to find the $$c_{1},c_{2}$$ since you've stated that you've already found the lambdas. The eigenvectors, v_1 v_2, you've already found.

Last edited: Aug 8, 2008
5. Aug 8, 2008

### HallsofIvy

Staff Emeritus
Another way of looking at it is this: since your matrix, I'll call it A, has two independent eigenvectors, it can be "diagonalized". That is, there exist an invertible matrix U such that
$$UAU^{-1}]= U\left[\begin{array}{cc}4 & -1 \\ -4 & 4\end{array}\right]U^{-1}= \left[\begin{array}{cc}6 & 0 \\ 0 & 2\end{array}\right]= D$$
and, of course, it follow from that that A= U-1DU. Now it should be easy to see that An= U-1DnU. For example, A2= (U-1DU)(U-1DU)= U-1D2U because the U-1 and U between the two "D"s cancel.

In fact, you can take U-1 to be the matrix having the eignvectors as columns:
$$\left[\begin{array}{cc}1 & 1 \\ -2 & 2\end{array}\right]$$
so that U is its inverse
$$\left[\begin{array}{cc}\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & \frac{1}{4}\end{array}\right]$$

You should be able to check that
$$\left[\begin{array}{cc}4 & -1 \\ -4 & 4\end{array}\right]=\left[\begin{array}{cc}1 & 1 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}6 & 0 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & \frac{1}{4}\end{array}\right]$$
So that
$$\left[\begin{array}{cc}4 & -1 \\ -4 & 4\end{array}\right]^n=\left[\begin{array}{cc}1 & 1 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}6^n & 0 \\ 0 & 2^n\end{array}\right] \left[\begin{array}{cc}\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & \frac{1}{4}\end{array}\right]$$
Mulitplying out the right side should give you the solution.

Last edited: Aug 8, 2008
6. Aug 8, 2008

### franky2727

so the information that i currently have is lamda1=2 lamda2=6 V1=(1,2)T and V2=(1,-2)T

What is r, K, and C and how do i go about finding them?

7. Aug 8, 2008

### franky2727

so A^n=U^-1 . D^n . A . U

where do my X0 and Y0 values fall into this i'm still confused but this is definatly closer to the way i remember doing this

8. Aug 8, 2008

### franky2727

im going to walk home now and get some food but ill check this again in about a hour thanks guys

9. Aug 8, 2008

### HallsofIvy

Staff Emeritus
?? No one has said anything about "r, K, and C". konthelion did say that the solution can be written in the form
$${r_{k}=c_{1} \lambda_{1}^k v_{1} + c_{2} \lambda_{2}^k v_{2}}$$
where c1 and c2 are constants to be determined by the "initial conditions (X0= Y0= 1), the two $\lambda$s are your eigenvalues, 2 and 6, and v1 and v2 are the corresponding eigenvalues.

Since you got $\lambda_1= 6$ with $v_1= \left[\begin{array}{c}1 \\ -2\end{array}\right]$ and $\lambda_2= 2$ with $v_2= \left[\begin{array}{c} 1 \\ 2\end{array}\right]$, You have
$$r_n= c_1 6^n \left[\begin{array}1 \\ -2\end{array}\right]+ c_2 2^n \left[\begin{array}{c}1 \\ 2\end{array}\right]$$
Taking n= 0,
$$r_n= c_1 \left[\begin{array}1 \\ -2\end{array}\right]+ c_2 \left[\begin{array}{c}1 \\ 2\end{array}\right]=\left[\begin{array}{c}c_1+ c_2 \\-2c_1+ 2c_2\end{array}\right]= \left[\begin{array}{c}1 \\ 1\end{array}\right]$$
so we get the two equations c1+ c2= 1 and -2c1+ 2c2= 1. You can easily solve those to get c1= 1/4 and c2= 3/4. Then
$$r_n= \frac{1}{4} 6^n \left[\begin{array}1 \\ -2\end{array}\right]+ \frac{3}{4} 2^n \left[\begin{array}{c}1 \\ 2\end{array}\right]$$
$$= 6^n\left[\begin{array}{c}\frac{1}{4} \\-\frac{1}{2}\end{array}\right]+ 2^n\left[\begin{array}{c}\frac{3}{4} \\ \frac{3}{2}\end{array}\right]$$
so that $x_n= (1/4)6^n+ (3/4)2^n$ and $y_n= (-1/2)6^n+ (1/2)2^n$.

Doing the calculation I suggested gives you a matrix. Now multiply that matrix by the vector (or column matrix)
$$\left[\begin{array}{c}x_0 \\ y_0 \end{array}\right]= \left[\begin{array}{c}1 \\ 1\end{array}\right]$$
That should give you the same solution as above.

Last edited: Aug 8, 2008
10. Aug 8, 2008

### franky2727

nice one i get it now thanks, having my tea right now but after that i'll go through this question and then have a look at the question i have without limits to see if i can do that and if not i'll put that one up as well, thanks for the help

11. Aug 9, 2008

### franky2727

i had a question yesterday entitled exam question help which 2 people were kind enough to work through with me to a point were i now understand how to do this question, as i stated on there yesterday i have another question that is similar but this time it is a bit more complex, could someone please take a look at this one and if possible try to present it in the same way that hallsofivey did yesterday as this was the way that i understood the question.

part (a) which i can do is find the solution to the two dimensional map

Xn+1=Xn -Yn
Yn+1=2Xn +4Yn

with X0=1 and Y0=1

giving me eivgenvalues of 2 and 3 with vectors of (1,-1)T and (1,-2)T respectivly

After which i do the question just the same as the one that was answered for me above and i end up with hopefully the correct answer of Xn=-2 . 3n+3 . 2n

and Yn=4.3n-3.2n

hopefully that bits right. then i have part b of the question which i dont know how to do which is

indicate how the second order map
Xn+1=4xn-3xn-1

can be expressed as a two dimensional map. by concidering the eigenvalues of the associated matrix show that Xn then theres a symbol that i have never seen before, it looks like a 8 on its side but part of the loop is missing on the right hand side then 3n for n large... thanks