Matrix multiplication: Communicative property.

  1. Matrix multiplication: Commutative property.

    Hello,

    First time poster.
    I have got a question about commutative property of matrix multiplication.
    Literature says that matrix multiplication is communicative only when the two matrices are diagonal.

    But, I have a situation with an 'Unitary' matrix. Actually it is the DFT matrix http://en.wikipedia.org/wiki/Discrete_Fourier_transform#The_unitary_DFT. And I multiply with a 'vector'.

    It seems that communicative property holds in this case. But I want to know what is the theoretical explanation, or the property as to why communicative property holds in this case.

    Thank you very much.
     
    Last edited: Jan 23, 2009
  2. jcsd
  3. You certainly misread that, because it is not true that the diagonal matrix is the only[/tex] type of matrices that are commutative, surely many matrices are not commutative, but some are.
    Take for example the set of 2x2 matrices of the form
    [tex]\begin{array}{cc}
    a & b \\
    -b & a
    \end{array}
    [/tex]
     
    Last edited: Jan 23, 2009
  4. Another silly counter-example:

    Matrices of the form:


    [tex]\begin{array}{cc}
    a & 0 \\
    0 & 0
    \end{array}[/tex]
     
  5. The silly counter-example is a diagonal matrix in which one of the entries on the diagonal happens to be 0.

    A clarification might be useful here. You have a matrix that is multiplied by a vector? That might be an (m x m) matrix multiplied by an (m x 1) vector. You can't multiply an (m x 1) with an (m x m).
     
  6. Heh, oops. The silly example is for some reason I was using an alternative definition for "diagonal" X-D
     
  7. Thats OK. Usually I'm the one saying oops.
     
  8. Thanks a lot for the replies guys.

    I am multiplying a (1xM) vector with the (MxM) unitary matrix.
    For the few random matrices I tried, it seems to be commutative. Atleast for the DFT matrix (Vandermond) I tried.
    I want to know if there is any property talking about this scenario.

    Thank you very much.
     
  9. Can your matrixes be diagonalized with the same similarity transformation?

    [tex]
    \begin{align*}
    U_1 &= V D_1 V^* \\
    U_2 &= V D_2 V^*
    \end{align*}
    [/tex]

    EDIT: fixed wrong lingo in question.
     
    Last edited: Jan 23, 2009
  10. D H

    Staff: Mentor

    You are using the term "commutative" incorrectly here. Given an operator * and operands a and b, a and b commute if a*b=b*a. If a is 1xM and b is MxM, the product a*b exists (and is 1xM) but b*a exists only if M=1. In other words, there is no way a 1xM and a MxM matrix can commute unless M=1 (i.e., if a and b are scalars).
     
  11. That's what I was getting at. That's why I was mentioning the dimensions of the matrices. Too often software packages bend the rules and treat vectors as both (1 x m) and (m x 1) matrices to fit the math. In that case a symmetric (m x m) matrix allows the operation to be commutative, i.e. A[i,j] = A[j,i].
     
  12. Sorry guys.

    Yes, if I am using say A is a (1xM) vector, it cannot be commutative.
    My mistake. Actually I am AxB and BxA', conjugate of A.
    So this is not commutation anymore?
    If this is the case, that means matrix multiplication properties cannot be applied, unless I put my vector entries inside a diagonal of a matrix?

    Thanks lot guys.
     
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