Product of Symmetric and Antisymmetric Matrix

Click For Summary
SUMMARY

The trace of the product of a symmetric matrix \( A \) and an antisymmetric matrix \( B \) is definitively zero. This conclusion is reached by utilizing the properties of traces, specifically that \( \text{Tr}(AB) = \text{Tr}(BA) \) and \( \text{Tr}(AB^T) = \text{Tr}(A(-B)) = -\text{Tr}(AB) \). By establishing that \( \text{Tr}(AB) = -\text{Tr}(AB) \), it follows that \( \text{Tr}(AB) = 0 \). The discussion also emphasizes the importance of understanding index notation and double summation in matrix operations.

PREREQUISITES
  • Understanding of symmetric and antisymmetric matrices
  • Familiarity with matrix trace properties
  • Proficiency in index notation and summation conventions
  • Basic knowledge of linear algebra concepts
NEXT STEPS
  • Study the properties of matrix traces in linear algebra
  • Learn about symmetric and antisymmetric matrices in depth
  • Practice index notation and double summation techniques
  • Explore examples of matrix products and their traces
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in advanced matrix theory and its applications.

ognik
Messages
626
Reaction score
2
Hi, I want to show that the Trace of the Product of a symetric Matrix (say A) and an antisymetric (B) Matrix is zero.
$So\: (AB)_{ij}=\sum_{k}^{}{a}_{ik}{b}_{kj} $
$and\: Tr(AB)=\sum_{i=j}^{}(AB)_{ij}=\sum_{i}^{}\sum_{k}^{}{a}_{ik}{b}_{ki} $
$because\:A\:is\:symetric, \: {a}_{ik}= {a}_{ki}\:so\:Tr(AB)=\sum_{i}^{}\sum_{k}^{}{a}_{ki}{b}_{ki}$
Here I am stuck - I want to say that because B is antisymetric, it's diagonal entries must be 0, but I am a bit weak with index notation, and especially with double summation signs - can't see how to show $b_{ki}$ is a diagonal element inside this summation ... I think :-)
 
Physics news on Phys.org
It is not necessary to use indices. Use the facts that $\operatorname{tr}A=\operatorname{tr}A^T$ and $\operatorname{tr}(AB)=\operatorname{tr}(BA)$.
 
Thanks Evgeny, I used Tr(ABT) = Tr(ATB)
Tr(ATB)=Tr(AB) and Tr(ABT)=Tr(A(-B))=-Tr(AB)
So Tr(AB)=-Tr(AB), therefore Tr(AB)=0
But if it can be done along the lines I tried with indexes, I'd really like to see that - I am looking for opportunities to practice Indexing :-)
Also I am still unsure what to do when I come across things like $\sum_{}^{}\sum_{}^{}$
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
5K
Replies
3
Views
898
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
12K
  • · Replies 2 ·
Replies
2
Views
2K