# What does it mean to be self adjoint with respect to an inner product?

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?

• Could you please use $\LaTeX$? It's unreadable. May 6, 2014 at 11:25
• "h j iB : Rn Rn ! R by h v j w iB = vtBw for all v;w 2 Rn" is completely and utterly illegible. Please use MathJax to typeset your formula, it really isn't hard to use.
– fgp
May 6, 2014 at 11:25
• And BTW, how do you define self-adjoint if not with respect to an inner product? The usual definition is $\langle Tx,y\rangle = \langle x,T^*y \rangle$, which of course is with respect to whatever inner product $\langle \cdot,\cdot \rangle$ refers to.
– fgp
May 6, 2014 at 11:27

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