# Matrix multiplication: Communicative property.

• m26k9
In summary, the conversation discusses the commutative property of matrix multiplication and how it applies to different types of matrices. One member mentions that the literature says matrix multiplication is communicative only when the two matrices are diagonal, but another member provides a counter-example and clarifies that there are other types of matrices that are commutative. The conversation then shifts to discussing a specific scenario where a unitary matrix is being multiplied by a vector and whether the commutative property still holds. It is clarified that the term "commutative" is being used incorrectly in this context and it is not possible for a 1xM and MxM matrix to commute unless M=1. The conversation concludes with the realization that matrix multiplication properties cannot
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.

Last edited by a moderator:
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
$$\begin{array}{cc} a & b \\ -b & a \end{array}$$

Last edited:
Another silly counter-example:

Matrices of the form:

$$\begin{array}{cc} a & 0 \\ 0 & 0 \end{array}$$

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).

hokie1 said:
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

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

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.

Thank you very much.

m26k9 said:
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.

Last edited:
m26k9 said:
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).

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.

## 1. What is the communicative property of matrix multiplication?

The communicative property of matrix multiplication states that changing the order of multiplication of two matrices does not change the result. In other words, for two matrices A and B, A * B = B * A.

## 2. Why is the communicative property important in matrix multiplication?

The communicative property is important because it allows for easier and more efficient calculations in matrix operations. It also helps in simplifying equations and solving systems of linear equations.

## 3. Does the communicative property hold true for all matrices?

No, the communicative property does not hold true for all matrices. It only applies to square matrices, where the number of rows is equal to the number of columns.

## 4. Can the communicative property be extended to more than two matrices?

No, the communicative property only applies to two matrices. It cannot be extended to more than two matrices.

## 5. How is the communicative property different from the associative property?

The communicative property states that the order of multiplication can be changed, whereas the associative property states that the grouping of matrices in multiplication can be changed without affecting the result. In other words, for three matrices A, B, and C, (A * B) * C = A * (B * C), but (A * B) may not necessarily be equal to (B * A).

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