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?
Best Answer
Hint: $A$ is self-adjoint w.r.t. $\langle\cdot,\cdot\rangle_B$ iff $$\color{green}{\langle Av,w\rangle_B}=v^TA^TBw=v^T\color{red}{(BA)^T}w=v^T\color{red}{BA}w=\color{green}{\langle v,Aw\rangle_B}\quad\forall v,w\in\mathbb{R}^n.$$