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Linear Algebra Dynamical Systems

  1. Dec 9, 2007 #1
    1. A = {[0.4 0 .2], [0.3 0.8 0.3], [0.3 0.2 0.5]}. The vector v1 = {[0.1], [0.6], [0.3]} is an eigenvector for A, and two eigenvalues are .5 and .2. Construct the solution of the Dynamical system x,k+1 = Ax,k that satisfies x,0 = (0, 0.3, 0.7)

    My attempt

    I tried to work this one out but I'm just lost as to where to begin, I think i have to start by finding all the eigenvalues and put them in a diagonal matrix and then put them in the equation: x,k = c1([tex]\lambda1[/tex])v1 + c2([tex]\lambda2[/tex])v2 Anyone got any ideas that could help me out?
  2. jcsd
  3. Dec 9, 2007 #2


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    By "solution to the dynamical system" you mean a formula for xk for all k?

    One problem I see is that you haven't copied the problem correctly! Your "A" is clearly supposed to be a 3 by 3 matrix but you have only two values for the first row!

    In any case you are given two of the eigenvalues and an eigenvector. Although, given two of the eigenvalues, it should be simple to find the third, I suspect that the eigenvector you are given corresponds to the third eigenvalue. Multiply A by <0.1, 0.6, 0.3> and see what multiple of <0.1, 0.6, 0.3> it is. If that multiple is not 0.5 or 0.2, then it is your third eigenvalue. If that multiple is either 0.5 or 0.2, then knowing that eigenvector doesn't help at all, but you should still be able to work out the characteristic polynomial for A, divide by (x- 0.5) and (x- 0.2) and have a single (x- a) left so that a is the third eigenvector. Since you haven't given A correctly here, I don't know which of those will work.
    Last edited: Dec 9, 2007
  4. Dec 9, 2007 #3
    no its copied correctly the values are [.4, 0, and .2]and yes a formula for x,i for all k
  5. Dec 9, 2007 #4


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    You'll need to find all of the eigenvectors and then how to express (0,0.3,0.7) as a sum of them. Then you can write A^k(x0) as a sum of the kth powers of the eigenvalues times the eigenvectors.
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