Prove $A^tB^t = (BA)^t$

Question

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!

Answer 1

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: $${}^{\mathrm t\!}(BA)={}^{\mathrm t\!}A {}\mkern2mu^{\mathrm t\!}B.$$

Answer 2

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 $A^\top B^\top=(BA)^\top$.