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Question on the trace of two matrices

  1. Oct 13, 2011 #1
    How does knowing that two matrices anticommute AB=-BA and that A^2=1 and B^2=1 help me to know how to find the trace of the matrices. I am supposed to show that their traces equal eachother which equals 0 but I am not sure exactly how the given information helps me determine the trace?
     
  2. jcsd
  3. Oct 13, 2011 #2
    do you mean A^2 = I?
     
  4. Oct 13, 2011 #3
    In general, trace(AB) = trace(BA). Hence from the given relation trace(AB) should be zero as +0 = -0.
    or do you mean trace(A) = trace(B) = 0?
     
    Last edited: Oct 13, 2011
  5. Oct 13, 2011 #4
    trace(A) = trace(B) = 0
    and the problem said A^2 =1
     
  6. Oct 13, 2011 #5

    Ben Niehoff

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    What size are the matrices?

    I can think of two ways to approach this problem:

    1. Assume the matrices are in upper-triangular form (this is always possible), and see what you can conclude about the diagonal elements, or

    2. Write out the invariant formula for the characteristic polynomial. The characteristic polynomial can always be written in terms of tr A, det A, and tr A^k for k up to n-1 (for an n x n matrix).

    If the matrices are 2x2, then the second method will be very easy. I'm not sure if what you're trying to prove is necessarily true if the matrices are larger.
     
  7. Oct 13, 2011 #6

    I don't know what size the matrices are I just know that they are two matrices A and B. I guess I'm still not understanding the connection of the trace to anticommuting and to the squares
     
  8. Oct 13, 2011 #7

    HallsofIvy

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    [itex](a+ b)^2= a^2+ ab+ ba+ b^2= 2i[/itex]

    (Why are my capital letters always turned into lower case?)
     
  9. Oct 13, 2011 #8
    Does i here refer to I (the identity matrix)?

    Maybe I'm being really dumb but where does that come from? Is it finding the eigenvalues of [itex](a+b)^2[/itex]
     
  10. Oct 13, 2011 #9

    I like Serena

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    Welcome to PF, evlyn! :smile:

    Yes, as HoI said, his text was turned into lower case.
    It should be:
    [tex](A + B)^2= A^2+ AB+ BA+ B^2= 2I[/tex]
    However, aside from the fact that it is a true statement, I do not know what he intended with it.

    As for your problem, if you look up "Minimal polynomial" (as Ben Niehoff suggested), you'll find that the minimal polynomial of A is a polynomial divisor of (A2 - I), due to the fact that A2 = I.
    In effect this means that the only eigenvalues A can have (and also B), are the roots of x2 - 1 = 0.
    That is, -1 and +1.

    Therefore the trace of A (and also of B) is the sum of n values of either -1 or +1.

    Furthermore, for a nxn matrix, the characteristic polynomial is [itex]x^n - tr(A) x^{n-1} + ... + (-1)^n \det(A)[/itex].
    In particular, for a 2x2 matrix, you can read off the trace and determinant of A from this.
     
    Last edited: Oct 13, 2011
  11. Oct 16, 2011 #10
    I don't understand how the minimal polynomial is helpful if I just have a general matrix A? I don't no anything about it except that AB = -BA and that A^2=1 and B^2=1
     
  12. Oct 16, 2011 #11

    I like Serena

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    Well, first off, I interpret your equation as A^2=I.

    From just that equation you can draw a number of conclusions about A.
    Let me enumerate them for you:
    1. A is a square nxn matrix (otherwise you could not multiply it with itself).
    2. A is its own inverse, that is, A = A^-1.
    3. More specifically: A is invertible.
    4. det(A^2)=1, implying det(A)=+1 or det(A)=-1.

    Now the minimal polynomial is helpful, since it tells us that:
    5. A has n eigenvalues, each of which must either be +1 or -1.


    Can you tell what the dimensions of B must be?


    Btw, I get the impression that the minimal polynomial is outside of the scope of your current class material.
    Is it?
    Perhaps we should zoom in on 2x2 matrices and leave the minimal polynomial out of it...
    And anyway, I'm not even sure myself if your statement holds true for matrices larger than 2x2.
     
    Last edited: Oct 16, 2011
  13. Oct 16, 2011 #12
    the dimensions of B must be the same as A


    It is most likely assumed that the general population in my class knows what the minimal polynomial is
     
  14. Oct 16, 2011 #13

    I like Serena

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    Yep!
    So A and B must both be nxn matrices.



    Probably not! :wink:
    It is advanced material in linear algebra.



    Let's suppose that [itex]A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}[/itex].

    Do you know what the determinant is?
    What the trace is?
    And what the characteristic polynomial is?
     
  15. Oct 16, 2011 #14
    The determinant would be ad-bc
    the trace would be a+d
    and the characteristic polynomial I am not entirely sure about ... (t-a)(t-d)-bc ...
    t^2-at-td+ab-bc
     
  16. Oct 16, 2011 #15

    I like Serena

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    Good! :)

    Do you know what the characteristic polynomial represents? What it is used for?
     
  17. Oct 16, 2011 #16
    It is used to find the eigenvalues of the matrix
     
  18. Oct 17, 2011 #17

    I like Serena

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    So what are you unsure about?
     
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