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