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Matrix trace

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data

    A square matrix A (of some size n x n) satisfies the condition A^2 - 8A + 15I = 0.

    (a) Show that this matrix is similar to a diagonal matrix.
    (b) Show that for every positive integer k >= 8 there exists a matrix A
    satisfying the above condition with tr(A) = k.

    2. Relevant equations

    A^2 - tr(A)A + det(A)I = 0

    3. The attempt at a solution

    I'm not entirely sure what to do but here's the attempt..

    I subtracted the given formula from the equation, obtaining
    (8 - tr(A))A + (15 - det(A))I = 0

    So either tr(A) = 8 or A = cI, where c = [15 - det(A)]/tr(A) - 8

    So since I've shown that A = cI, c being a constant, is this enough to show that A is similar to a diagonal matrix?

    For (b), I'm completely lost... If it means use A^2 - kA + 15I = 0
    then surely the result will follow straight from the last part? And presumably having k<8 will result in something impossible?
     
  2. jcsd
  3. Aug 22, 2011 #2
    Do you think you could write the original equation as (A-3I)(A-8I).
    If you could justify writing it out like that, and you looked at the determinant of that expression, what would you find out?
     
  4. Aug 22, 2011 #3

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    Interesting problem.
    I do not have a solution yet.
    Since I do not know your relevant equation, I decided to analyze it.

    If I take a 1x1 matrix A = (a), I get:

    (a)2 - tr[(a)] (a) + det[(a)] I = (a2) - a . (a) + a . (1) = (a)

    However, this is not equal to zero!?

    For a 2x2 matrix it does check out, always.

    For a 3x3 matrix it is not generally true, although there are solutions.

    Where did you get your equation?
     
  5. Aug 22, 2011 #4

    micromass

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    Do you know about "minimal polynomials" and the Cayley-Hamilton theorem??

    Can you prove (using this) that the characteristic equation of A is exactly [itex]x^2-8x+15[/itex]?
     
  6. Aug 22, 2011 #5

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    @micromass: I did not know the Cayley-Hamilton theorem yet, nor "minimal polynomials".
    Thanks for the reference. :smile:
    However, as far as I can tell now, it is only applicable to this problem for n=2, as is your characteristic equation.
    Am I missing something?

    @OP: I suspect these theorems are beyond the level you are at. Am I wrong?
     
  7. Aug 22, 2011 #6

    hunt_mat

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    How about this, let P be an invertible matrix, the:
    [tex]
    P^{-1}(A^{2}-8A+15I)P=0\Rightarrow P^{-1}APP^{-1}AP-8P^{-1}AP+15I=0\Rightarrow (P^{-1}AP)^{2}-8P^{-1}AP+15I=0
    [/tex]
    All you need to do now is show that A has a full set of eigenvectors and you can see that A is similar to a diagonal matrix.

    To examine the eigenvalues, write [itex]Ax=\lambda x[/itex] and examine [itex]A^{2}x[/itex] and [itex]Ix[/itex], this should leave you with three equations which can manipulated to give you an equation for [itex]\lambda[/itex]
     
    Last edited: Aug 22, 2011
  8. Aug 22, 2011 #7

    micromass

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    No, you're not missing anything :smile: I failed to see that it had to be an arbitrary size nxn :frown:
     
  9. Aug 22, 2011 #8

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    Errr... "All you need to do now is show that A has a full set of eigenvectors"?
    You make it sound almost easy...
    Any suggestions how to do that?
    Because I've got no clue. :frown:
     
  10. Aug 22, 2011 #9

    hunt_mat

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    :smile: I see what you're saying, I provided some suggestions to examine the eigenvalues. I will have another think about it.
     
  11. Aug 22, 2011 #10

    micromass

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    Well, you can do it with minimal polynomials (even for arbitrary sizes).
    Here's a sketch:
    First show that the minimal polynomail of A divides [itex]x^2-8x+15[/itex] (this is true by definition).
    Since the roots of the minimal polynomial are exactly the eigenvectors, one can show that 3 and 5 are the only possible eigenvectors.

    So if n>2, then trace(A)>8, and a technique as in the OP shows that the matrix is diagonalizable. So only n=2 remains.
     
  12. Aug 22, 2011 #11
    The equation I supplied at the start may only be for a 2x2 matrix, as I found it in a solution my lecturer had for a different problem on a 2x2 matrix, and incorrectly assumed it was valid for an nxn matrix.

    And yes, I know the Cayley-Hamilton theorem, it is on my course, but I haven;t been exposed to it enough to fully understand it
     
  13. Aug 22, 2011 #12

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    Yeah, well, I may have been exposed to it as well, but if I have been, I've long forgotten it! :wink:
    Lack of enough exposure I guess.
     
  14. Aug 22, 2011 #13
    Or lack of going to lectures. :wink:

    So, from what I've gathered, since the minimal polynomial of A is x^2 - 8x + 15, and the roots of this are 3 and 5, then the eigenvalues are 3 and 5.

    Since it has eigenvalues it presumambly has a jordan normal form and is thus similar to a diagonal matrix?
     
  15. Aug 22, 2011 #14

    micromass

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    Not necessarily. It can happen that A only has 3 as eigenvalue, for example.
     
  16. Aug 22, 2011 #15

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    Hold on, I get (after some research) that 3 and 5 are the only possible eigenvalues, but how would you know that A has n eigenvalues?

    Btw, for n=2 we can solve the system, finding either the 2 eigenvalues {3, 5}, meaning A is diagonizable, or finding 2 specific solutions for A, which are diagonal matrices.
     
  17. Aug 22, 2011 #16

    micromass

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    It has n eigenvalues up to multiplicity :smile:
    Hmm, the case n=1 and n=2 is easy. But the higher order cases still elude me...
     
  18. Aug 22, 2011 #17

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    Now that I think about it, any nxn matrix has n eigenvalues, although they may be complex.
    Since 3 and 5 are the only possible eigenvalues, the possibly complex eigenvalues would have to be either 3 or 5 as well!
     
  19. Aug 22, 2011 #18

    micromass

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    Ah, you have seen Jordan normal forms?? Good!!

    No, a Jordan normal form is in general not a diagonal matrix. But in this case it is!

    Remember that the sizes of the largest Jordan block is the multiplicity of the eigenvalue in the minimal polynomial. And here the multiplicity is always one, thus the Jordan blocks all have size 1.
     
  20. Aug 22, 2011 #19

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    A Jordan normal form is not necessarily similar to a diagonal matrix.
    If the Jordan normal form contains a block, it isn't.

    Edit: aarrrrgggh
     
  21. Aug 22, 2011 #20

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    Huh? :uhh:

    If an eigenvalue has a multiplicity of m, that only means it occurs m times in the Jordan normal form in blocks of size 1.
    However, if the minimal polynomial would have 2 complex (conjugate) roots, that would correspond to a block of size 2.
     
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