How to prove that $(AB)^T=B^TA^T$ with property $v⋅Aw=A^Tv⋅w$ How to prove that $(AB)^T=B^TA^T$ with property $v\cdot Aw=A^Tv\cdot w$. I know how $(AB)^T=B^TA^T$ is normally derived, I am wondering how to prove it with $v\cdot Aw=A^Tv\cdot w$. Thank you!
 A: We have $(AB)^T v\cdot w=v\cdot ABw=A^Tv\cdot Bw=B^TA^T v\cdot w$ for all $v,\,w$, so $(AB)^T=B^TA^T$.
A: You are given that
$$
v\cdot Aw=A^Tv\cdot w\qquad\text{for all $v$, $w$ and $A$}
$$
(where $v$ and $w$ are vectors and $A$ is a square matrix of suitable dimension).
You are also given that the scalar product is nondegenerate, so $v\cdot w=0$ for every $w$ implies $v=0$.
Now apply the property to $v$, $Bw$ and $A$:
$$
v\cdot ABw=A^Tv\cdot Bw
$$
Apply the property to $A^Tv$, $w$ and $B$:
$$
v\cdot ABw=A^Tv\cdot Bw=B^TA^Tv\cdot w
$$
You can also apply the property to $v$, $w$ and $AB$:
$$
v\cdot ABw=(AB)^Tv\cdot w
$$
Hence
$$
(AB)^Tv\cdot w=B^TA^Tv\cdot w
$$
and, by linearity,
$$
\bigl((AB)^Tv-B^TA^Tv)\cdot w=0
$$
for every $v$ and $w$. By non-degeneracy,
$$
(AB)^Tv-B^TA^Tv=0
$$
for all $v$, which means
$$
\bigl((AB)^T-B^TA^T\bigr)v=0
$$
for all $v$. Therefore $(AB)^T-B^TA^T=0$.

This seems like a waste of time, because proving directly the identity is much simpler.
However, this can be generalized. Let $V$ be a finite dimensional inner product space (not necessarily $\mathbb{R}^n$ with the standard scalar product). For a linear operator $T$ on $V$ we can define its adjoint $T^*$ as the unique linear operator such that, for all $v,w\in V$,
$$
T^*v\cdot w=v\cdot Tw
$$
(proving it exists is a nice exercise). Then, with the same technique as above, we can prove that for two operators $T$ and $U$,
$$
(TU)^*=U^*T^*
$$
