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.
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]
Another silly counter-example: Matrices of the form: [tex]\begin{array}{cc} a & 0 \\ 0 & 0 \end{array}[/tex]
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
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.
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.
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).
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].
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.