Prove $A^tB^t = (BA)^t$ Just going over some old homework problems for my test tomorrow. One of the questions was the prove $A^tB^t = (BA)^t$ and at the time I was really unsure of my answer and wrote the following:


*

*My Answer:


We can write the $ij^{th}$ entry of $(BA)$ as $(BA)_{ij} = \sum_{k=1}^m{b_{ik}a_{kj}}$. 
Coincidentally, $(A^tB^t)_{ij}=\sum_{k=1}^n(A^t)_{ik}(B^t)_{kj} = \sum_{k=1}^na_{ki}b_{jk}$
Is this a logical answer? It just seemed too... loose for me.
Thank you!
 A: Your idea is essentially correct. For a complete proof note that
\begin{align*}
[A^\top B^\top]_{ij}
&= \sum_{k=1}^m [A^\top]_{ik}[B^\top]_{kj} \\
&= \sum_{k=1}^m [A]_{ki}[B]_{jk} \\
&= \sum_{k=1}^m [B]_{jk}[A]_{ki} \\
&= [BA]_{ji} \\
&= [(BA)^\top]_{ij}
\end{align*}
Hence $A^\top B^\top=(BA)^\top$.
A: I will give details on the coordinates-free proof:
$A$ is the matrix  of a linear map $f\colon E\to F$ in some bases of vector spaces $E$ and $F$, $B$ is likewise the matrix of a linear map $g\colon F\to G$, after a basis in $G$ has been chosen,  and $BA$ is the matrix of $g\circ f:E\to G$.
Now the  transposed map $\,{}^{\mathrm t\!}f:F^\ast\to E^\ast$ is defined by  $\,{}^{\mathrm t\!}f(\varphi)= \varphi\circ f$. It has matrix $\,{}^{\mathrm t\!}A$ in the dual bases of $E^\ast$ and  $F^\ast$.
If $ \varphi\in G^\ast$:
$$\,{}^{\mathrm t\!}(g\circ f)(\varphi)=\varphi\circ(g\circ f)=(\varphi\circ g)\circ f={}^{\mathrm t\!}g( \varphi)\circ f= {}^{\mathrm t\!}f\bigl({}^{\mathrm t\!}g( \varphi)\bigr),$$
which proves  $\,{}^{\mathrm t\!}(g\circ f)= {}^{\mathrm t\!}f\circ {}^{\mathrm t\!}g $, hence the matrices of both sides are equal:
$${}^{\mathrm t\!}(BA)={}^{\mathrm t\!}A {}\mkern2mu^{\mathrm t\!}B.$$
