Proving that the composition of orthogonal transformations is orthogonal in a finite dimensional inner product space $V$.

Question

Given $S$,$T$ orthogonal transformations in an inner product space $V$ over $\mathbb R$ such that $ \dim(V) < \infty$.

Also given that $v \in V$ | ($S$$\circ$$T) (v) = S(T(v))$

I need to prove that $S$$\circ$T is also an orthogonal transformation.

My answer is: Based on the given info that S and T orthogonal transformation

1)$\langle T(v),T(w) \rangle = \langle v,w \rangle $ and $\langle S(v),S(w) \rangle = \langle v,w\rangle $
2)$ \langle ST(v),ST(w) \rangle = \langle S(T(v)),S(T(w)) \rangle $ can I say that from here $ \langle S(v),S(w) \rangle$ and that equals to $\langle v,w \rangle$

Answer 1

No you cannot. You must "peel" the expression $\langle S(T(v)),S(T(w)) \rangle$ from outside, i.e. use that $\langle S(x),S(y) \rangle =\langle x,y\rangle$ (applied to $x=T(v),y=T(w)$), to first get rid of $S.$

Note that the hypothesis $\dim(V)<\infty$ is superfluous, but could be used to convert this to a matricial proof.

Answer 2

$\langle ST(v),ST(w) \rangle =\langle S(T(v)),S(T(w)) \rangle=\langle T(v),T(w) \rangle =\langle v,w \rangle$