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Recall that the transpose of a matrix $A$ is denoted by $A^t$. Let $K$ be a field and denote the $m$x$n$-matrices over $K$ by $M_{m×n}(K)$.

Prove that $(AB)^t = B^tA^t$ for all $A ∈ Mm×k(K)$, $B ∈ M_{k×n}(K)$. You should use the formula for the $(i,j)$-entry of the product of two matrices, and you should use that the $(k, l)$-entry of the transpose of a matrix $C$ is the $(l, k)$-entry of $C$.

I know I would begin by stating $AB$ is an $m×n$- matrix then $(AB)^t$ must be too, but don't know how to continue. What is the next step?

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First we note that $AB$ is an $m × n$-matrix and thus $(AB)^t$ is an $n × m$-matrix. Similarly, $B^t$ is an $n×k$-matrix and $A^t$ is an $k×m$-matrix, so $B^tA^t$ is an $n×m$-matrix. It remains to prove that the $(i, j)$-entry of $(AB)^t$ is equal to the $(i, j)$-entry of $B^tA^t$. For any matrix $C$, we write $C_{st}$ for the $(s, t)$-entry. Then $C_{st} = (C ^t)_{ts}$ for any matrix $C$. Using the formula for the product of two matrices, we see that the $(j, i)$-entry of $AB$ is $\sum_{s=1}^kA_{js}B_{si}$ Hence, the $(i, j)$-entry of $(AB)^t$ is also

$ \sum_{s=1}^kA_{js}B_{si} $

The $(i, j)$-entry of $B^tA^t$ is

$\sum_{s=1}^k(B^t)_{is}(A^t)_{sj}=\sum_{s=1}^kB_{is}A_{js}=\sum_{s=1}^kA_{js}B_{is}$

Hence the $(i,j)$-entry of $(AB)^t$ equals the $(i,j)$-entry of $B^tA^t$ proving $(AB)^t=B^tA^t$

How is that?