symmetric matrix multiplication

Question

I have three matrices $A$, $H$, and $B$. $A$, $H$, and $B$ are $n$ by $n$ matrices, and $H$ is a symmetric matrix. Is $A^THB$ equal to $B^THA$?

Answer 1

Oh okay. I see you have edited your question. In general, there is no reason for $A^THB$ to equal $B^THA$. These two matrices are the transposes of each other, so you will have $A^THB=B^THA$ if and only if $A^THB$ is symmetric. You can find several examples where $A^THB$ is not symmetric, even if $H$ is a symmetric matrix other than the identity. For example, let $$A=\left( \begin{array}{cc} 1 & 0\\ 1 & 1 \end{array} \right),\hspace{.5cm}B=\left( \begin{array}{cc} 0 & 1\\ 2 & 1 \end{array} \right),\hspace{.5cm}H=\left( \begin{array}{cc} 1 & 2\\ 2 & 1 \end{array} \right).$$ Then $$A^THB=\left( \begin{array}{cc} 6 & 6\\ 2 & 3 \end{array} \right)\neq\left( \begin{array}{cc} 6 & 2\\ 6 & 3 \end{array} \right)=B^THA.$$

It is really not hard to find examples where $A^THB\neq B^THA$. I didn't give any effort for my construction. I just picked some $2\times 2$ matrices off the top of my head, and they turned out to have the property that $A^THB\neq B^THA$.