Prove: symmetric positive matrix multiplied by skew symmetric matrix equals 0 My teacher gave me this task as preparation for the exam but I'm stuck and not sure if it's true anymore.
 A: I think your teacher means Frobenius product.
In the context of tensor analysis (e.g. widely used in mechanics, think about $\int \boldsymbol{\sigma}:\boldsymbol{\epsilon}\,\mathrm{d}\Omega$, if you know the weak form of elastostatics), it is a natural inner product for 2nd order tensors, whose coordinates can be represented in matrices.
The result is straightforward
$$A:B = \operatorname{tr}(A^T B) =a_{ij}b_{ij} = -a_{ji}b_{ji} = - A:B \Rightarrow A:B = 0.$$
A: Say $J$ is skew symmetric and $R$ is symmetric and invertible (a strictly positive definite matrix is invertible) .
Then 
$$
J R = ( R^{-1} R ) J R = R^{-1} ( R J R ) = 0
$$
where the last equality follows by properties of skew-symmetric matrices.
A: Let J and R be skew symmetric and symmetric matrices. Index notation of these matrices are;
$J_{ij}R_{ij}$=-$J_{ji} R_{ij}$     (1),
since R is symmetric matrix and can be written as $R_{ji}$=$R_{ij}$
Equation 1 becomes,
$J_{ij}$$R_{ij}$=$-J_{ji}$$R_{ji}$, swapping the indices on RHS will give,
$J_{ij}$$R_{ij}$=-$J_{ij}$$R_{ij}$ finally we obtain
$JR=-JR$  $\Rightarrow$  $2JR=0$ and $JR=0$
A: Let A be a symmetric matrix and B be a skew-symmetric matrix. i.e., $A_{ij} = A_{ji}$ and $B_{ij} = -B_{ji}$.
Define: C = AB
Prove that Tr(C) = Tr(AB) = 0
$C_{ik} = A_{ij}B_{jk}$
Using the properties above, the following is also true:
$C_{ik} = A_{ij}B_{jk} = A_{ji}(-B_{kj})$
and
$C_{ik} = A_{ij}B_{jk} = -A_{ji}B_{kj}$.
If we multiply through by $\delta_{ik}$:
$\delta_{ik}C_{ik} = \delta_{ik}A_{ij}B_{jk} = -\delta_{ik}A_{ji}B_{kj}$
which yields
$C_{ii} = A_{ij}B_{ji} = -A_{ji}B_{ij}$
$C_{ii} = C_{ii} = -C_{jj}$
Note that $C_{ii}$ and $C_{jj}$ are the same and are both equal to Tr(C).
Tr(C) = Tr(C) = -Tr(C)
For a scalar number to be equal to its negative, it must be zero.
$\therefore$ Tr(C) = 0
