Let $A$ and $B$ be $d\times d$, complex matrices such that $Ax\cdot x>0$ and $Bx\cdot x>0$ for every nonzero $x\in\mathbb C^d$, let $\left\langle x,y\right\rangle\_A=Ax\cdot y$, and let $\left\langle x,y\right\rangle\_B=Bx\cdot y$.

I showed that $\left\langle\cdot,\cdot\right\rangle\_A$ and $\left\langle\cdot,\cdot\right\rangle\_B$ are inner products on $\mathbb C^d$.

Let $\left\|x\right\|\_A=\sqrt{\left\langle x,x\right\rangle\_A}$ and let $\left\|x\right\|\_B=\sqrt{\left\langle x,x\right\rangle\_B}$.

I want to show that $\left\|\cdot\right\|\_A$ and $\left\|\cdot\right\|\_B$ are equivalent.

I believe that the right approach is to show $\left\|\cdot\right\|\_A$ is equivalent to $\left\|\cdot\right\|\_I$, where $I$ is the identity matrix, but I have not had much success.