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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?

mdp
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Mark
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    What is the transpose of a linear map? – Travis Willse Oct 23 '14 at 11:18
  • This is a good idea, because you'll be proving a more fundamental statement. Just write out the definitions of the maps applied to an arbitrary functional and see that they agree! This is fiddly to write down, but ultimately nothing very complicated is happening. If you have already done this but got stuck somewhere, it might help to say what you did. – mdp Oct 23 '14 at 11:21

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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.

mdp
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    If I let $(g ◦f)^t(a)=a ◦(g ◦ f)=(a ◦g) ◦f = f^t(g^t(a))=(f^t ◦g^t)(a)$ where a is in W*, is this sufficient? – Mark Oct 23 '14 at 11:37
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    Yes - as long as you can prove equality at some level, it's fine. I was conservative in the suggestion and went all the way down to evaluation on $U$, but as you point out, just playing with composition rules lets you prove equality after evaluation on $W^*$. – mdp Oct 23 '14 at 13:53
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Define $f^t:f^t(\alpha)=\alpha\circ f$ and think of it as a pull-back of $\alpha$. Let $U\overset{f}{\to}V\overset{g}{\to}W$. Then we can pull back any linear functional $\alpha\in W^*$ in the reverse order to act on $U$: $(g\circ f)^t(\alpha)=\alpha\circ (g\circ f)=(\alpha\circ g)\circ f=f^t(\alpha\circ g)=f^t(g^t(\alpha))=f^t\circ g^t(\alpha).$

rych
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    The OP had concluded that in a comment to the previous answer more than a month before you posted the same thing. – PinkyWay Mar 15 '20 at 09:54