Matrix multiplication: Communicative property.

[m26k9]

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.

 

[wsalem]

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 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}<br /> a &amp; b \\<br /> -b &amp; a<br /> \end{array}<br />

 

[Tac-Tics]

Another silly counter-example:

Matrices of the form:


\begin{array}{cc}<br /> a &amp; 0 \\<br /> 0 &amp; 0<br /> \end{array}

 

[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).

 

[Tac-Tics]

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

 

[hokie1]

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

 

[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. 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.

 

[Peeter]

For the few random matrices I tried, it seems to be commutative.

Can your matrixes be diagonalized with the same similarity transformation?

<br /> \begin{align*}<br /> U_1 &amp;= V D_1 V^* \\<br /> U_2 &amp;= V D_2 V^*<br /> \end{align*}<br />

EDIT: fixed wrong lingo in question.

 

[D H]

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).

 

[hokie1]

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].

 

[m26k9]

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.