The discussion revolves around the expression ##(A_{ik}B_{kj})_{mm} = (A_{ki}B_{jk})_{mm}## in index notation, focusing on the properties of matrix multiplication and the implications of symmetry and anti-symmetry in the context of traces of matrices. Participants explore the notation and the conditions under which the equality holds.
Participants generally express uncertainty about the symmetry assumption, and while some agree on the properties of traces, there is no consensus on the implications of the anti-symmetric part or the conditions under which the original equality holds.
There are unresolved questions regarding the notation used and the assumptions about the matrices A and B, particularly concerning their symmetric and anti-symmetric components.
It is true for the anti-symmetric part also because the sign change cancels.hellomrrobot said: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 said:It is true for the anti-symmetric part also because the sign change cancels.
Yes, I think so. A and B can be decomposed into symmetric and anti-symmetric parts.hellomrrobot said:anti-symmetric?
It would help if you would explain your notation.hellomrrobot said: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?
gill1109 said: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).
Thank you! You are right.stevendaryl said:Well, I would say that the right hand side is trace(BA), but that is always equal to trace(AB).