Alternative proof of a transpose property

Question

I am asked to prove; $$(AB)^T=B^TA^T$$ although it is very simple to prove it by the straight forward way, in the exercise I am asked to prove it without using subscripts and sums, directly from the following property of inner product of real vectors:$$\langle A\textbf{x},\textbf{y}\rangle =\langle \textbf{x},A^T\textbf{y}\rangle$$

Where $A$ is an $m\times n$ real matrix, $x \in \mathbb{R}^n$,$y \in \mathbb{R}^m$

I don't know how to approach this problem, any suggestions?

Answer 1

You could do the following:

$$\langle x,A^\intercal B^\intercal y\rangle=\langle Ax,B^\intercal y\rangle = \langle BAx, y\rangle = \langle x,(BA)^\intercal y\rangle$$

Do you know how to show that this implies $A^\intercal B^\intercal=(BA)^\intercal$?

Answer 2

Use nondegeneracy of inner product. I.e if for every vector $x\in\mathbb{R}^n$ $$<x,y_1>=<x,y_2>$$ holds, then $y_1=y_2.$ Fix $y\in\mathbb{R^m}$ and from what you have quoted you have that for every $x$ $$<ABx,y>=<Bx,A^Ty>=<x,B^TA^Ty>$$ and $$<ABx,y>=<x,(AB)^Ty>.$$ Hence $B^TA^Ty=(AB)^Ty.$ $y$ was arbitrary, hence $B^TA^T=(AB)^T.$