Proving (A_{ik}B_{kj})_{mm} = (A_{ki}B_{jk})_{mm}: Index Notation Question

[hellomrrobot]

I am having trouble showing that $(A_{ik}B_{kj})_{mm} = (A_{ki}B_{jk})_{mm}$ Wouldn't the right side end up having a different outcome? Or can we assume its symmetric?

 

[Mentz114]

I am having trouble showing that $(A_{ik}B_{kj})_{mm} = (A_{ki}B_{jk})_{mm}$Wouldn't the right side end up having a different outcome? Or can we assume its symmetric?

It is true for the anti-symmetric part also because the sign change cancels.

 

[hellomrrobot]

It is true for the anti-symmetric part also because the sign change cancels.

anti-symmetric?

 

[Mentz114]

anti-symmetric?

Yes, I think so. A and B can be decomposed into symmetric and anti-symmetric parts.
For the antisymmetric part
$A_{ik}B_{kj}=(-A_{ki})(-B_{jk})$

[Edit]

Is there a 'raised' index here ?

${A_i}^k B_{kj}$

 

[gill1109]

I am having trouble showing that $(A_{ik}B_{kj})_{mm} = (A_{ki}B_{jk})_{mm}$Wouldn't the right side end up having a different outcome? Or can we assume its symmetric?

It would help if you would explain your notation.

I suppose that you are using Einstein summation convention. And that when you write (A_{ik}B_{kj}) you mean, the matrix whose i,j element is sum_k (A_{ik}B_{kj}). In other words, the ij element of the matrix AB. Now you want to sum over m the m,m elements of that matrix, so you are talking about trace(AB).

Similarly on the right hand side, replace i and j both by m and m, sum over m and sum over k, and you see that what you have written is just trace(AB).

 

[stevendaryl]

It would help if you would explain your notation.

I suppose that you are using Einstein summation convention. And that when you write (A_{ik}B_{kj}) you mean, the matrix whose i,j element is sum_k (A_{ik}B_{kj}). In other words, the ij element of the matrix AB. Now you want to sum over m the m,m elements of that matrix, so you are talking about trace(AB).

Similarly on the right hand side, replace i and j both by m and m, sum over m and sum over k, and you see that what you have written is just trace(AB).

Well, I would say that the right hand side is trace(BA), but that is always equal to trace(AB).

 

[gill1109]

Well, I would say that the right hand side is trace(BA), but that is always equal to trace(AB).

Thank you! You are right.