How to demonstrate $\operatorname{tr}(AB)=0$ when $A$ is symmetric and $B$ is antisymmetric For a square matrix $A$ of order $n$, let $\operatorname{tr}(A)$ be the sum of the diagonal entries.
$\mbox{tr}(A)=\sum_{i=1}^{n} \textrm{a}_\textrm{ii}$

Assume that $A$ is symmetric and $B$ is antisymmetric, how to demonstrate that $\operatorname{tr}(AB)=0$?

 A: Hint. Use these basic properties of the trace operator: for any $n\times n$ square matrices $M$ and $N$ then 
$$\operatorname{tr}(-M)=-\operatorname{tr}(M),\quad \operatorname{tr}(M)=\operatorname(M^t)\quad\text{and}\quad \operatorname{tr}(MN)=\operatorname{tr}(NM).$$
Note that in your case $(AB)^t=B^tA^t=-BA$.
A: $\operatorname{tr}(AB)=\operatorname{tr}(AB)^t=\operatorname{tr}(B^tA^t)$
$A$ is symmetric so $A^t=A$, B is antisymmetric $B^t=-B$
$$\operatorname{tr}(AB)=\operatorname{tr}(B^tA^t)=\operatorname{tr}(-BA)=-\operatorname{tr}(BA)=-\operatorname{tr}(AB)$$
$2\operatorname{tr}(AB)=0 \iff \operatorname{tr}(AB)=0$ 
A: $$\text{Tr}(AB)=\sum_{i=1}^n(AB)_{ii}$$
$$=\sum_{i=1}^n \sum_{j=1}^n A_{ij} B_{ji}$$
$$=\sum_{i=1}^n \sum_{j=1}^n A_{ji} (-B_{ij})$$
$$=-\sum_{j=1}^n \sum_{i=1}^n A_{ji} B_{ij}$$
$$=-\sum_{j=1}^n (AB)_{jj}$$
$$=-\text{Tr}(AB)$$
So
$$\text{Tr}(AB)=0$$
A: $A$ symmetric means $a_{ji} = a_{ij}$ for all $i,j$, while $B$ antisymmetric means $b_{ii} = 0$ and $b_{ji} = - b_{ij}$. 
Now check that for any $n\times n$ matrices $A$, $B$ the trace of the product is 
$$S= \sum_{ij} a_{ij} b_{ji}$$
The sum can be broken into
$$S_1= \sum_{i<j} a_{ij}b_{ji}\\
S_2 = \sum_{i>j} a_{ij}b_{ji}\\
S_3 = \sum_{i=j} a_{ij}b_{ji}$$
Clearly $S_3=0$. Moreover, we have
$$S_2 = \sum_{i<j} a_{ji}b_{ij}= \sum_{i<j} a_{ij}(-b_{ji})= - S_1$$
Summing up we get $S = S_1 + S_2 + S_3 = S_1 - S_1 + 0 = 0$.
