Statement: Suppose that $B\in M_n(\mathbb{R})$ is a symmetric matrix such that $v^TBv > 0$ for all nonzero vectors $v\in\mathbb{R}^n$. Here $v^T$ denotes the transpose of $v$. Define the map $\langle\cdot,\cdot\rangle_B: \mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ by $\langle v,w\rangle_B=v^TBw$ for all $v,w\in\mathbb{R}^n$. You may assume without proof that $\langle\cdot,\cdot\rangle_B$ is an inner product on $\mathbb{R}^n$, which we will call the $B$-inner product. Throughout this question we consider $\mathbb{R}^n$ as an inner product space with respect to the $B$-inner product. Let $A\in M_n(\mathbb{R})$, which we consider as a linear map $A:\mathbb{R}^n\to\mathbb{R}^n$.
Question: Prove that if $BA$ is symmetric then $A$ is self-adjoint with respect to the $B$-inner product. With this question I am unsure what it means to be self-adjoint with respect to an inner product?