TITLE: Transpose of product of matrices
QUESTION [13 upvotes]: How do you prove the following fact about the transpose of a product of matrices? Also can you give some intuition as to why it is so.
$(AB)^T = B^TA^T$

REPLY [2 votes]: I marked this as community wiki since it so close to Saketh Malyala's answer.

For the intuition/background, please read this site answer.
We will now prove the assertion.
For any matrix $C$ let $\text{Row}(C,i)$ denote the $i^\text{th}$ row of $C$ represented in a natural way as vector.
For any matrix $C$ let $\text{Col}(C,j)$ denote the $j^\text{th}$ column of $C$ represented in a natural way as vector.
The $(i,j)^\text{th}$ entry of $AB$ is equal to $\langle \text{Row}(A,i), \text{Col}(B,j)\rangle$
The $(j,i)^\text{th}$ entry of $B^tA^t$ is equal to $\langle \text{Row}(B^t,j), \text{Col}(A^t,i)\rangle$
But $\text{Row}(B^t,j) = \text{Col}(B,j)$ and $\text{Col}(A^t,i) = \text{Row}(A,i)$, so indeed,
$$ {(AB)}^t = B^t A^t$$