Stuck on proving $(AB)^t=B^tA^t$

Question

Let $A\in \mathcal{M}_{m,n}(\mathbb{C}), B\in \mathcal{M}_{n,p}(\mathbb{C})$. I need to show that $(AB)^t=B^tA^t$.
I know that this is a very well-known property, but I am a bit stuck. So, I considered that $A=(a_{ij})_{1\le i\le m \\ 1\le j\le n}$ and $B=(b_{jk})_{1\le j \le n \\ 1\le k\le p}$ and then I took $X=AB=(x_{ik})_{1\le i\le m \\ 1\le k \le p}$, where $\displaystyle x_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk}$.
However, I don't know how to write the elements of $X^t$ now. I mean, I know that $X^t=(x_{ki})_{1\le k \le p \\1\le i \le m}$, but I don't know how to write $x_{ki}$ explicitely.
P.S.: I am aware that there are other questions on this website where this was proved, but none of them explains this step that puzzles me.
EDIT: I tried to add some more work, let me know what the problem is:
$(AB)^t=(x_{ki})_{1\le k \le p \\ 1\le i\le m}$, $\displaystyle x_{ki}=\sum_{j=1}^{n}a_{kj}b_{ji}$ (I just switched $i$ and $k$ in $x_{ik}$'s formula) $B^tA^t=(y_{ki})_{1\le k \le p \\ 1 \le i \le m}$, $\displaystyle y_{ki}=\sum_{j=1}^n b_{kj}\cdot a_{ji}$

Answer 1

Just compute each product and notice it matches..

$$ (AB)_{ij}= \sum_{k=1}^n a_{ik} b_{kj}\Rightarrow (AB)^T_{ij} = \sum_{k=1}^n a_{jk}b_{ki} $$

$$ (B^T A^T)_{ij} = \sum_{k=1}^n b^T_{ik} a^T_{kj} = \sum_{k=1}^nb_{ki} a_{jk} $$

Answer 2

Short answer:

The trick that you need is

$$a_{ij}=a^T_{ji}$$

where $a^T$ denotes an element of the transposed matrix $A^T$.

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Long answer:

An element in a matrix product is the dot product of a row of the first with a column of the second.

If you swap the matrices and transpose them, you compute exactly the same dot products (to a harmless swap of the arguments).

If you prefer "equations",

$$\sum_{k}a_{ik}b_{kj}=\sum_{k}b^T_{jk}a^T_{ki}.$$

Answer 3

Denote $c_{ij}$ the coefficients of $AB\enspace (1\le i\le m,\:1\le j\le p)$, $\:a_{ik}\enspace (1\le i\le m,\:1\le k\le n)$ and $b_{kj}\enspace (1\le k\le n,\:1\le j\le b)$ those of $A$ and $B$. Finally denote with a $'$ the coefficients of the transposed matrices:

• ${}^{\mathrm t\kern-2mu}(\kern-2mu AB)$ is a $p{\times}n$ matrix and its generic coefficient is $c'_{ij}=c_{ji}$
• By definition, $$ c'_{ij}=c_{ji}=\sum_{k+1}^n a_{jk}b_{ki}=\sum_{k=1}^n a'_{kj}b'_{ik}=\sum_{k=1}^n b'_{ik}a'_{kj}=\bigl(\kern1mu^{\mathrm t\!}B\:^{\mathrm t\kern-6mu}A\bigr)_{ij}.$$