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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!

David South
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  • It's fine, though it would probably be better to take the coordinate-free approach: For $X: V \to W$, the map $X^t:W^* \to V^*$ is defined by $X^t(f) = f(X)$; and so $(BA)^t(f) = f(BA) = B^t (f(A)) = A^t B^t f$. – anomaly Jun 10 '15 at 20:06
  • Could you explain the equality "$f(BA) = B^t(f(A))$ – David South Jun 10 '15 at 20:09
  • Just write out the definition of $B^t$. – anomaly Jun 10 '15 at 20:11
  • I would add explicitly what is ${}^{\mathrm t}(BA){ij},$ with respect to $(BA){kl}$. I would even start from from there. – Bernard Jun 10 '15 at 20:24

2 Answers2

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

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

Bernard
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