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Tensor notation-Determinants

  1. May 30, 2008 #1
    [SOLVED] Tensor notation-Determinants

    I'm trying to learn the basics of Tensor calculus using a free online book (Introduction to Tensor Calculus and Continuum Mechanics), and I got stuck on this question (Part 2 in book, after non-math introduction).

    link to part 2, questions (mine #19) at end:

    1. The problem statement, all variables and given/known data
    Let A and B denote 3x3 matrices with elements [tex]A_{ij}[/tex] and [tex]B_{ij}[/tex]
    respectively. Show that if C = AB is a matrix product, then det(C) = det(A)*det(B)
    where det = determinant.

    Hint: use the result from example 1.1-9

    2. Relevant equations
    det(A) = [tex]e_{ijk}A_{1i}A_{2j}A_{3k}[/tex]

    3. The attempt at a solution
    The matrix multiplication of A*B =C, in indical notation, is
    [tex]C_{ij}=A_{im}B_{mj}[/tex] (I think) where the first subscript in A,B and C is the row number and the second subscript the column of the matrix.

    Then, plugging into 'relevant equation' above, we get
    det(C) = [tex]e_{ijk}C_{1i}C_{2j}C_{3k}[/tex]
    det(C) = [tex]e_{ijk}(A_{1m}B_{mi})(A_{2n}B_{nj})(A_{3x}B_{xk})[/tex]

    Then, I compare this to just multiplying det(A)*det(B)
    det(A) * det(B) = [tex](e_{ijk}A_{1i}A_{2j}A_{3k})(e_{rst}B_{1r}B_{2s}B_{3t})[/tex]

    However, from here, I can't seem to make the connection between the two:cry:
    I would try to expand it, but I don't think this is the way to do it, since it would get rid of the advantage of indical notation. Help would be appreciated! Thxs!

    PS: on a side note, whenever you preview a post, the template is pasted again to the end of the post, which is annoying. Maybe someone can fix this...
  2. jcsd
  3. May 30, 2008 #2
    I believe you need to prove this in two cases. If B is not invertible, then det(B)=0. There is a non-trivial solution. Thus AB is not invertible.

    Else, suppose A and B are invertible/nonsingular, then there are elementary matrices, A=Ej, E_j-1,.... E_1

    Now, det(AB)= det(E_j)*det(E_j-1)...(E_1) * det(B).
    Then A and B can be expressed as a product of elementary matrices.

    Edit:Since C=AB, then det(C)=det(AB). Hope that helps.
    Last edited: May 30, 2008
  4. May 30, 2008 #3


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    Maybe I'm reading the intent incorrectly, but I think the problem specifically wants him to prove it using indicial notation.
  5. May 31, 2008 #4
    Oh sorry.

    Let C=AB. By indicial notation, C_ij=A_ik B_kj

    Since A is 3x3 matrix, then det(A)= A_11(A_22...) - A_12(A21..) + A_13(A_21...) by method of solving for determinants where n>2
    Then det(A)= epsilon_ijk A_i1 A_j2 A_k3 = epsilon_ijk A_1i A_2j A_3k
    Therefore, epsilon_rst det(A) = epsilon_ijk A_ir A_js A_kt

    We know that det(C)=det(AB)=epsilon_ijk C_i1 C_j2 C_k3
    =...= epsilon_rst B_r1 B_s2 B_t3 det(A)
  6. May 31, 2008 #5
    I can't believe i didn't see it! Thanks guys!:smile:
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