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Eigenvalue(s) for a matrix

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the eigenvalues of A, when A =[3,1,1; 0,5,0; -2,0,1]


    2. Relevant equations



    3. The attempt at a solution

    I took the det(A-[itex]\lambda[/itex]I)=0 and found the characteristic polynomial to be:

    -[itex]\lambda[/itex][itex]^{3}[/itex]+15[itex]\lambda[/itex][itex]^{2}[/itex]-73[itex]\lambda[/itex]+115

    I couldn't figure out whole roots to the equation and wolframalpha gave me
    [itex]\lambda[/itex]=5, 5-√2, 5+√2

    Does this seem like the right answer? I have checked and double checked the determinant. Would I still be able to find a basis for the eigenspace?
     
  2. jcsd
  3. Aug 1, 2012 #2

    sharks

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    For matrix:
    $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 1\end{bmatrix}$$
    Here is the characteristic equation that i get:
    ##-λ^3+9λ^2-25λ+25=0##
     
    Last edited: Aug 1, 2012
  4. Aug 1, 2012 #3
    Because the second column is [0, 5, 0], the polynomial should look like (5 - x)Q(x), where Q(x) is a second-degree polynomial. So it should be very easy to get its roots. Do it.
     
  5. Aug 1, 2012 #4
    Sorry the matrix was $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 7\end{bmatrix}$$
    not 2,0,1 in the last row. That was my mistake.
     
  6. Aug 1, 2012 #5
    Yes, those are the eigenvalues.
     
  7. Aug 1, 2012 #6

    Mark44

    Staff: Mentor

    The second row is [0, 5, 0].
     
  8. Aug 1, 2012 #7
    This is not clear from the original notation. Anyway, that does not matter for determinants so my argument holds.
     
  9. Aug 2, 2012 #8

    vela

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    Expand the determinant along the second row.
     
  10. Aug 2, 2012 #9

    sharks

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    Well, the original notation used is actually very common and can be recognized as a standard, since it is also used in popular software like MATLAB.
     
  11. Aug 2, 2012 #10
    Let's put it this way: not clear for me, I always seem to forget the conventions for rows vs columns in a linear transcription. This may stem from the fact that even when working with paper and pencil, I am about equally likely to write down a matrix column after column and row after row. Let's hope this thread will leave a dent in my state of confusion :)
     
  12. Aug 2, 2012 #11

    sharks

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    Your characteristic equation from post #1 is correct. Now, you just need to solve it. Usually, you would start by trial and error to find one factor of the polynomial (you already know the easy one is (λ-5)) and then do long division to obtain a quadratic in the quotient, which you would then solve using the quadratic formula.
     
  13. Aug 2, 2012 #12
    The point I made in #3 is that none of this is required. By tackling the determinant along [0, 5, 0], one gets the polynomial directly in the factored form.
     
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