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Linear Algebra - Matrix with given eigenvalues

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Come up with a 2 x 2 matrix with 2 and 1 as the eigenvalues. All the entries must be positive.
    Then, find a 3 x 3 matrix with 1, 2, 3 as eigenvalues.

    3. The attempt at a solution
    I found the characteristic equation for the 2x2 would be λ2 - 3λ + 2 = 0. But then I couldn't get a matrix with positive entries to work for that.
     
    Last edited: Apr 4, 2012
  2. jcsd
  3. Apr 3, 2012 #2

    Dick

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    Pick a diagonal matrix.
     
  4. Apr 3, 2012 #3
    Does that count for the entries being positive though?
     
  5. Apr 3, 2012 #4

    Dick

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    Not really, no. Sorry. Better give this more thought than I gave this response.
     
  6. Apr 3, 2012 #5
    thanks though!
     
  7. Apr 4, 2012 #6

    micromass

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    The 2x2-case is not so difficult. Remember (or prove) that the characteristic polynomail of a 2x2-matrix A is

    [tex]\lambda^2-tr(A)\lambda+det(A)[/tex]

    By the way, I think your characteristic polynomial is wrong.
     
  8. Apr 4, 2012 #7

    HallsofIvy

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    ??? Why do the diagonal matrices
    [tex]\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}[/tex]
    and
    [tex]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}[/tex]
    NOT count as "all entries postive"?
     
  9. Apr 4, 2012 #8

    micromass

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    He probably doesn't consider 0 to be positive.
     
  10. Apr 4, 2012 #9

    HallsofIvy

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    But it is much easier to claim that 0 is positive!:tongue:

    Thanks.
     
  11. Apr 4, 2012 #10
    Oh, I had typed 3 instead of 2 for the characteristic polynomial. I ended up looking at this from a Hermitian matrix point of view.
    And then I got the matrix:
    0 i +1
    i-1 3
    And I did get the right eigenvalues from that. Does that work?
     
  12. Apr 4, 2012 #11

    micromass

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    You still have 0 as an entry, you don't want that.
     
  13. Apr 4, 2012 #12
    Yeah, I didn't realize that at first. :/
     
  14. Apr 4, 2012 #13

    Dick

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    I don't think i+1 would be considered a positive number either. Stick to real entries. Your diagonal entries need to sum to 3, and their product should be greater than 2. Do you see why?
     
  15. Apr 4, 2012 #14
    Yes. Theoretically, I know what it should do. I just can't actually find the right values to do it.
     
  16. Apr 4, 2012 #15

    Dick

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    Call one diagonal entry x. Then the other one must be 3-x. Can you find a positive value of x that makes x*(3-x)>2? Graph it.
     
  17. Apr 4, 2012 #16
    Well, any value of x between 1 and 2 (like 1.1) work.
     
  18. Apr 4, 2012 #17

    Dick

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    Ok, so you just need to fill in the rest of the matrix.
     
  19. Apr 4, 2012 #18
    but if I set x to be 1.1, my matrix would be
    1.1 __
    __ 1.9

    And those two spaces have to be equivalent to 1.1*1.9 - 2, right?
    because no matter what values I try, when the eigenvalues are getting closer to 1 and two, the matrix is just getting closer to the matrix of:
    1 0
    0 2
     
  20. Apr 4, 2012 #19

    Dick

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    The two spaces multiplied together have to give you 1.1*1.9 - 2. How about putting one blank to be 1 and the other to be 1.1*1.9 - 2? The eigenvalues should work out to EXACTLY 1 and 2. Try it with x=3/2.
     
  21. Apr 4, 2012 #20
    I had just figured out that
    1.5 0.5
    0.5 1.5
    worked out! :)
     
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