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Alternative method to the matrix problem?

  1. Oct 24, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    Find all real matrices,X, of the form

    [x 0]
    [y z]

    such that X2+3X+2I=0

    (I is the identity matrix)


    2. Relevant equations



    3. The attempt at a solution

    The only method I know is to work out X2 and then 3X, add to 2I, then equate the corresponding entries with the zero matrix. Have 3 equations with three unknowns and solve. But is there any other method in which I can use to shorten the amount of writing I will have to do?
     
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  3. Oct 24, 2008 #2

    morphism

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    If X^2+3X+2I=0, then the minimal polynomial of X divides t^2+3t+2=(t+1)(t+2), and take it from here.
     
  4. Oct 24, 2008 #3

    rock.freak667

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    I was thinking to do something like that, but the identity matrix in it confused me.
    Now I should re-write the equation as

    X^2++X+2I^2=(X+I)(X+2I)=0 and go from here right?
     
  5. Oct 24, 2008 #4

    morphism

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    Well, it depends on what you mean by "go from here"! If you're going to say something like "Then X=-I or X=-2I", then you'll be off the mark. Do you know what a minimal polynomial is?
     
  6. Oct 24, 2008 #5

    rock.freak667

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    Yes, that is what I thought to do.... No idea what a minimal polynomial is.
     
  7. Oct 24, 2008 #6

    morphism

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    OK, do you know what eigenvalues are, then?

    By the way, for matrices, AB=0 doesn't imply that A=0 or B=0. Try to come up with an example of this - and make sure you understand why this happens. In fact, you can have a nonzero matrix A such that A^2=0.
     
  8. Oct 24, 2008 #7

    rock.freak667

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    Yes,I know what eigenvalues are, and the characteristic polynomial. I am aware the AB=0 doesn't always mean that A=0 or b=0...I always forget this in matrices.
     
  9. Oct 24, 2008 #8

    morphism

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    Then I guess a shortcut would be to notice that the eigenvalues of a triangular matrix will lie on its diagonal. And because X^2+3X+2I=0, the eigenvalues of X will satisfy the equation t^2+3t+2=0. So this'll give us why x and z will be. To find y you'll just have to do it the long way.

    Well come to think of it, this isn't really a shortcut at all.
     
  10. Oct 24, 2008 #9

    rock.freak667

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    I guess I will have to stick with my initial method.
    My math course, doesn't include the topic of e.vectors/values but since I did Further math I just happen to know it.
     
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