I am trying to find an alternative proof that $(AB)^t=B^tA^t$. I think it is possible to do by showing that:
$(g \circ f)^t=f^t \circ g^t$
where f,g are linear maps between vector spaces. I am unsure how to show this, can anyone point me in the right direction?
Best Answer
Let $f\colon U\to V$ and $g\colon V\to W$. If you want to check the maps $(g\circ f)^t$ and $f^t\circ g^t$ are equal, you need to check they agree on an abritrary input $\alpha\in W^*$. So you need to check
$$(g\circ f)^t(\alpha)=(f^t\circ g^t)(\alpha).$$
As pointed out in the comments, this equality can already be proved directly.
Alternatively, both sides of this equation are also functions, now in $U^*$, so you could check they agree on an arbitrary input $u\in U$. So you need to check
$$((g\circ f)^t(\alpha))(u)=((f^t\circ g^t)(\alpha))(u).$$
Now this is just a case of writing out the definitions and seeing that they agree.