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Find the Trace and Determinant

  1. Dec 16, 2011 #1
    T:V -> V is linear.
    V is finite vectorspace of dimension m^2.
    T(M) = AMB where M is an mXm matrix and A, B are two fixed mXm matrices.
    I want to find the trace and determinant of this transformation.
    In the case where B is the indentity, I can show that the trace is m*tr(A) and the determinant is m*det(A). This is so because the matrix of this linear map can be written as an m^2Xm^2 matrix with a bunch of As on the diagonals. Do i proceed in the same way when B is not the identity? It looks complicated.
    (Solving a few easy examples led me to believe that the trace and determinant is the same as in the special case...or maybe i chose bad matrices...)

    Please HELP!
     
    Last edited: Dec 16, 2011
  2. jcsd
  3. Dec 16, 2011 #2
    Uh...Is the formula for the trace (double summation over i,j from 1 to m) a(i,i)*b(i,j) ?
    Is this supposed to be just a computational question? Am I missing something?
     
  4. Dec 16, 2011 #3

    Mark44

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    The trace of a square matrix is the sum of the entries on the main diagonal.
    [tex]Tr(A) = \sum_{i = 1}^n a_{ii}[/tex]

    The trace of a matrix and its determinant are generally different. For example, Tr(I2) = 2 and det(I2) = 1.
     
  5. Dec 16, 2011 #4
    Yeah, I know what the trace of any square matrix is. When I said 'formula for the trace', I meant for the transformation I was talking about; T(M) = AMB A,B being fixed matrices, M being the input matrix and a(i,j) b(i,j) being the entries of the matrices A and B.
    Is my formula correct?
     
    Last edited: Dec 16, 2011
  6. Dec 16, 2011 #5

    vela

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    Is there some condition on M that you haven't mentioned? How can the trace and determinant not depend on the matrix M? For example, if M=0, the trace and determinant would be 0, but your expressions would be non-zero if tr(A)≠0.

     
  7. Dec 16, 2011 #6
    M is the vector being transformed! I'm talking about the trace and determinant of the transformation T! M is a variable coming from the set of mXm matrices. T transforms M by multiplying by two matrices A and B. I am not talking about det/trace of the product AMB! I am asking about the trace and determinant of the m^2 X m^2 matrix representing the transformation!

    Vela, your question is equivalent to asking how the determinant and trace of a matrix is affected by the vector it multiplies.

    I am angry! Grr!
     
  8. Dec 16, 2011 #7

    vela

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    Oops!

    I used Mathematica to explicitly calculate the matrix for 3x3 matrices, and it looks like you should get that the trace is equal to tr(A)tr(B).
     
    Last edited: Dec 16, 2011
  9. Dec 16, 2011 #8
    oh! Hmm, i'll check my work again. How do I go about finding the determinant though?
    Should I be looking at eigenvalues?
     
  10. Dec 16, 2011 #9
    about the trace, both our formulae seem to be wrong...
     
  11. Dec 16, 2011 #10

    vela

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    I'm not sure, but looking at eigenvalues sounds like a reasonable thing to do.

    I'm only inferring what the result should be by working out special cases. You wrote above that if B=I, you thought the determinant should be m*det(A). Wouldn't it be det(A)m if you have a bunch of A's on the diagonal?
     
  12. Dec 16, 2011 #11
    yes. det(A)^m. Sorry.
    Even in the case where B is not the identity, the matrix of the transformation seems to be made up of identical(up to scalar multiplication) block matrices. But they are not restricted to the diagonal.

    The formula for the trace looks like double summation a(i,i)*b(j,j)
     
  13. Dec 16, 2011 #12

    vela

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    This is what I see for the 2x2 case. I chose for my basis the matrices
    \begin{align*}
    M_1 &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \\
    M_2 &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\
    M_3 &= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \\
    M_4 &= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
    \end{align*}In this basis, I found T represented by the matrix
    \begin{pmatrix}
    a_{11} b_{11} & a_{11} b_{21} & a_{12} b_{11} & a_{12} b_{21} \\
    a_{11} b_{12} & a_{11} b_{22} & a_{12} b_{12} & a_{12} b_{22} \\
    a_{21} b_{11} & a_{21} b_{21} & a_{22} b_{11} & a_{22} b_{21} \\
    a_{21} b_{12} & a_{21} b_{22} & a_{22} b_{12} & a_{22} b_{22}
    \end{pmatrix}So the trace appears to be tr(A)tr(B). Is this what you get?
     
  14. Dec 16, 2011 #13
    Ah yes! It is tr(a)tr(b)!
    But there has got to be a smarter way to do this question....
     
  15. Dec 16, 2011 #14
    in the case where A,B are diagonalizable with eigenvalues xi and yi (i from 1 to m) you can show that the eigenvalues of the transformation will be of the form xi * yj (i,j from 1 to m). This can be shown by putting eigenvectors corresponding to a particular eigenvalue of B transpose as the rows of M and eigenvectors corresponding to a particular eigenvalue of A as the columns of M. (Note: the eigenvalues of B and B transpose are the same)

    I don't know what to do if the eigenvectors of A or B don't exist.
     
    Last edited: Dec 17, 2011
  16. Dec 17, 2011 #15

    Dick

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    Extend the m^2xm^2 matrix expansion Vela gave you to higher dimensions. It splits up into mxm block matrices. Each block is an single element of A multiplied by the B matrix. It's an outer product (or a tensor product or Kronecker product). You can compute the trace by computing the trace of the blocks.
     
    Last edited: Dec 17, 2011
  17. Dec 17, 2011 #16
    ok thanks!
     
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