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Matrix multiplication: Communicative property. 
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#1
Jan2309, 12:25 AM

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. 


#2
Jan2309, 04:20 AM

P: 56

Take for example the set of 2x2 matrices of the form [tex]\begin{array}{cc} a & b \\ b & a \end{array} [/tex] 


#3
Jan2309, 08:44 AM

P: 810

Another silly counterexample:
Matrices of the form: [tex]\begin{array}{cc} a & 0 \\ 0 & 0 \end{array}[/tex] 


#4
Jan2309, 08:53 AM

P: 42

Matrix multiplication: Communicative property.
The silly counterexample 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). 


#5
Jan2309, 09:14 AM

P: 810




#6
Jan2309, 09:25 AM

P: 42

Thats OK. Usually I'm the one saying oops.



#7
Jan2309, 11:45 AM

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. 


#8
Jan2309, 02:07 PM

P: 295

[tex] \begin{align*} U_1 &= V D_1 V^* \\ U_2 &= V D_2 V^* \end{align*} [/tex] EDIT: fixed wrong lingo in question. 


#9
Jan2309, 02:58 PM

Mentor
P: 15,167




#10
Jan2309, 08:59 PM

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].



#11
Jan2409, 02:18 AM

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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