# Matrix multiplication: Communicative property.

 P: 9 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/Discret...he_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.
P: 56
 Literature says that matrix multiplication is communicative only when the two matrices are diagonal.
You certainly misread that, because it is not true that the diagonal matrix is the [b]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
$$\begin{array}{cc} a & b \\ -b & a \end{array}$$
 P: 810 Another silly counter-example: Matrices of the form: $$\begin{array}{cc} a & 0 \\ 0 & 0 \end{array}$$
 P: 42 Matrix multiplication: Communicative property. 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).
P: 810
 Quote by hokie1 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).
Heh, oops. The silly example is for some reason I was using an alternative definition for "diagonal" X-D
 P: 42 Thats OK. Usually I'm the one saying oops.
 P: 9 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.
P: 295
 Quote by m26k9 For the few random matrices I tried, it seems to be commutative.
Can your matrixes be diagonalized with the same similarity transformation?

\begin{align*} U_1 &= V D_1 V^* \\ U_2 &= V D_2 V^* \end{align*}

EDIT: fixed wrong lingo in question.
Mentor
P: 15,055
 Quote by m26k9 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.
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).
 P: 42 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].
 P: 9 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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