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Similarity - trace and determinant

  1. Jul 26, 2005 #1
    What exactly is the relationship between the trace/determinant of two matrices with regards to similarity. I always thought that if the trace was the same, then there is a possibility that the matrices are similar and if the determinant was the same, then the matrices are similar. On a recent exam we were given three matrices

    A
    1 0 1
    2 3 5
    0 2 -6

    B
    -4 3 4
    0 1 2
    0 0 1

    C
    0 0 2
    0 4 1
    3 5 -2

    One of these matrices is similar to A.

    I found det(a) = det (c) but trace (a) is not equal to trace (c). Det(B) is not equal to det (a) but trace (a) = Trace (b). Do I have my facts wrong?
     
  2. jcsd
  3. Jul 26, 2005 #2

    AKG

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    If two matrices are similar, then they have the same determinant, although it isn't necessarily true that if they have the same determinant, that they're similar. Anyways, this tells you that the matrix similar to A must have the same determinant as A. This is because if A is similar to B, then A = UBU-1 for some U, and:

    det(A) = det(UBU-1) = det(U)det(B)det(U-1) = det(U)det(B)det(U)-1 = det(B)

    I can't remember any special properties about traces and similarities, but the above should be enough to answer your question anyways.
     
  4. Jul 26, 2005 #3

    matt grime

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    If A and B are similar then tr(A)=tr(B) and det(A)=det(B) but absolutely no reverse deduction can be made.

    [tex]\left( \begin{array}{cc} 1&1\\ 0&1 \end{array}\right)[/tex] and


    [tex]\left( \begin{array}{cc} 1&0\\ 0&1 \end{array}\right)[/tex]

    have the same determinant and the same trace (and indeed the same characteristic poly) but are not similar, indeed trace and det are data contained in the char poly and that is the same for all similar matrices. Simillarity is hard to classify simply but can be done with jordan blocks.


    For AKG, matrices satisfy the cyclic trace property, and is perhaps weaker than being an invariant of the char poly in that it needs no manipulation other than by simply looking at the entries of the matrices: tr(ABC)=tr(BCA)=tr(CAB), and that tr(ABC)=/=tr(BAC). In fact it is even simpler than that now i come to think of it: tr(AB)=tr(BA)
     
    Last edited: Jul 26, 2005
  5. Jul 26, 2005 #4

    AKG

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    Could you expand on that, matt? What is the cyclic trace property? I'm not sure what is meant by the stuff after that too:

    "and is perhaps weaker than being an invariant of the char poly in that it needs no manipulation other than by simply looking at the entries of the matrices"

    EDIT: Oh wait, I think I see. The cyclic trace property is that the trace is invariant even if you cycle the matrices A, B, and C around, so as you said:

    Tr(ABC) = Tr(BCA) = TR(CAB)

    I still don't get the stuff in quotes above though.
     
  6. Jul 26, 2005 #5

    lurflurf

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    A scalar invarient of a matrix (with respect to similarity) is a function
    f:nxn matricies->scalar
    such that
    f(AS)=f(SA)
    for all nxn matrices A and all invertible nxn matrices S
    So tr is an invariant
    two matrices A & B are similar if
    f(A)=f(B) for all invariants
    It can be shown that a nxn matrix has n independent invariants thus one could chose their favorite set of n invariants and test those. Independent means that the invarians do not satisfy any relations.
    common such sets include
    -eigenvalues of A
    -coefficients of the characteristic polinomial of A (which det(A) and tr(A) are)
    -tr(A^k) k=1,...,nn

    So to summerize (~ will mean similar)
    A~B->tr(A)=tr(B)
    but tr(A)=tr(B) does not imply similarity
    n-1 more independent invariants need also be equal.
     
  7. Jul 27, 2005 #6

    matt grime

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    Well, we can shor Tr(AB)=tr(BA) by just working out what they are. usign summation convention they are

    A_{ij}B_{ji} and B_{ij}A_{ji} which are obviously equal.

    hence tr(AB)=tr(BA) is true for elementary reasons, ie we nned not talk abuot characteristic polys or sets of eigenvectors with multiplicity. which was how i first thought of provign the statement until i remembered you don't need to be a smart arse all the time. that's all i meant when i talked about elementary methods; i was talking to myself.
     
  8. Jul 28, 2005 #7

    lurflurf

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    These are the characteristic polynomials of these matricies.
    A^3+2A^2-31A^1+24A^0=0
    B^3+2B^2-7B^1+4B^0=0
    C^3-2C^2-19C^1+24C^0=0
    Since none of these have the same coefficients no pair of these is similar.
     
  9. Jul 28, 2005 #8

    Galileo

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    In view of the previous posts, the answer seems clear. A is similar to A.
     
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