MHB Product of Symmetric and Antisymmetric Matrix

[ognik]

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 :-)

 

[Evgeny.Makarov]

It is not necessary to use indices. Use the facts that $\operatorname{tr}A=\operatorname{tr}A^T$ and $\operatorname{tr}(AB)=\operatorname{tr}(BA)$.

 

[ognik]

Thanks Evgeny, I used Tr(AB^T) = Tr(A^TB)
Tr(A^TB)=Tr(AB) and Tr(AB^T)=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 $\sum_{}^{}\sum_{}^{}$