- #1
ognik
- 643
- 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 :-)
$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 :-)