I have a square matrix $B_{nxn}$ and a vector $A_{nx1}$ where $B$ and $A$ are independent with each other. I solved $Cov(A,BA)$ and obtained $Var(A)E(B)$ as follows,
$Cov(A,BA)=E(A^2)E(B)-E(A)(E(A)E(R))$
$=[E(A^2)-(E(A))^2]E(B)$
$=Var(A)E(B)$
$Var(A)$ will result in a scalar value. But, $B$ is a matrix. So, the result just not justify enough for $Cov(A,BA)$ which is a scalar value.
Please be kind enough to provide any instruction on the mistake I have made.
If $Var(A)E(B)$ is not correct, how can $Cov(A,BA)$ be solved and what is the answer for that?
Since we are dealing with vectors and matrices, we must be careful with operations like squares and so on. The calculation would go like this I think:
\begin{eqnarray*}Cov(A,BA)&=&E[(A-E[A])(BA-E[BA])^T]\\ &=&E[(A-E[A])(B(A-E[A]))^T]\\ &=&E[(A-E[A])(A-E[A])^TB^T]\\ &=&Cov(A)E[B]^T\end{eqnarray*}
Here $Cov(A)$ is the covariance matrix of the vector $A$, which corresponds to $Var(A)$. So the result is basically what you expected, but we need to respect matrix rules.