Transpositions wrt the energy and usual norm

Question

Let $^\dagger$ denote transposition w.r.t. the usual norm $(\cdot,\cdot)$ and $^*$ denote transposition w.r.t. the energy norm $(\cdot,\cdot)_A$ defined for symmetric $A$: $(u,v)_A=(Au,v)=(u,Av)$.

If $A$ is symmetric positive definite, how can I show that $(BA)^*=B^\dagger A^\dagger$?

My solution: $(BA)^*$ is the matrix s.t. $((BA)Au,v)=(Au,(BA)^*v)\forall u,v.$ We know $(Au,v)=(u,Av)\Rightarrow (AAu,v)=(Au,Av)\Rightarrow (BAAu,v)=(Au,B^TAv) \forall u,v$. So $B^TA=(BA)^*$. Does this make sense?

Answer 1

We say that $C = D^*$ if for any vectors $u,v$, we have $(Cu,v)_A = (u,Dv)_A$.

With that in mind: for any vectors $u,v$ in our Hilbert space, we have $$ ((BA)u,v)_A = (ABAu,v) = (BAu,A^\dagger v) = (Au,B^\dagger A^\dagger v) = (u,B^\dagger A^\dagger v)_A $$ thus we have reached the desired conclusion.