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Can someone please check this work (eigenvalues)

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Let

    [tex]
    A = \left[ \begin{array}{cc} -6 & 0.25 \\ 7 & -3 \end{array} \right]
    [/tex]

    Find an invertible S and a diagonal D such that [tex]S^{-1}AS=D[/tex]



    2. Relevant equations
    I basically have the question answered, just ONE problem.


    3. The attempt at a solution

    My answer is:

    [tex]
    S = \left[ \begin{array}{cc} 1 & -1 \\ 14 & 2 \end{array} \right]
    [/tex]

    [tex]
    D = \left[ \begin{array}{cc} -40 & 0 \\ 0 & -104 \end{array} \right]
    [/tex]

    I asked the professor what was wrong and he said "check the eigenvalues (main diagonal of D). The eigenvectors look ok, but their order has to match that of the eigenvalues.

    I've checked over and over, the math works out(S^-1 A S = D) , but I can't see what's wrong with the order. It looks perfectly fine for me.

    All I need to figure out is what the correct order is. Please help ><
     
    Last edited: Dec 3, 2009
  2. jcsd
  3. Dec 3, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    The math does not work out 'fine'. S^(-1)AS=[[-5/2,0],[0,-13/2]] as I work it out. I'll agree with your professor. The eigenvalues are wrong, but the eigenvectors are right. How did you do that?
     
  4. Dec 3, 2009 #3
    Well D is what I got from multiplying inverse of S, which was the eigen vectors I got multiplied by A, and then again by S, so...

    [tex]

    \left[ \begin{array}{cc} 2 & 1 \\ -14 & 1 \end{array} \right]
    \left[ \begin{array}{cc} -6 & .25 \\ 7 & -3 \end{array} \right]
    \left[ \begin{array}{cc} 1 & -1 \\ 14 & 2 \end{array} \right]

    [/tex]

    Which gave me my D which I guess is obviously wrong.


    [tex]

    D = \left[ \begin{array}{cc} -40 & 0 \\ 0 & -104 \end{array} \right]

    [/tex]

    My eigenvalues I got were 6.5 and 2.5 from the characteristic polynomial. Is that what's supposed to go into D's main diagonal or something? I thought I had to multiply it out like that.
     
  5. Dec 3, 2009 #4

    Dick

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    Homework Helper

    That helps. Your inverse of S is wrong. Try multiplying (inverse S)*S. You don't get the identity. Do you? Did you forget to divide by a determinant? And yes, the diagonal of D should be the eigenvalues.
     
    Last edited: Dec 3, 2009
  6. Dec 3, 2009 #5
    Oh dear, I did forget about that! Thanks! I'll try again.
     
  7. Dec 3, 2009 #6
    Okay so,

    [tex]

    S = \left[ \begin{array}{cc} 1 & -1 \\ 14 & 2 \end{array} \right]

    [/tex]

    [tex]

    S = \left[ \begin{array}{cc} -2.5 & 0 \\ 0 & -6.5 \end{array} \right]

    [/tex]

    Is this perfectly correct? I only get one more submission for this, so I need to make sure this is 100% correct. I'm only unsure because the prof said that "the eigenvectors look okay, but order has to match eigenvalues). Is that matched?
     
  8. Dec 3, 2009 #7

    Dick

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    You mean D= on the second line, right? And there's more than one S that will diagonalize A. So I'm not promising your homework system will accept it. But that looks ok to me. At least you have the eigenvalues on the diagonal.
     
  9. Dec 3, 2009 #8
    Whoops yeah, I meant D, and yeah I found that there were other values of S that would diagonalize A, some of which were just switching the signs of the elements around. I think you're right, the answer should be correct now.
     
  10. Dec 3, 2009 #9

    Dick

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    Homework Helper

    Well, you'll have to trust that the homework scanner will accept any correct answer. (You could also multiply S by 2 and multiply (inverse S) by 1/2.) If not, you have good grounds for appeal.
     
  11. Dec 3, 2009 #10
    Yeah the answer is correct now. All this new material made me forget the earlier stuff :(

    Thanks a lot, appreciate the help.
     
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