Why, intuitively, is the order reversed when taking the transpose of the product?

It is well known that for invertible matrices $A,B$ of the same size we have $$(AB)^{-1}=B^{-1}A^{-1} $$ and a nice way for me to remember this is the following sentence:

The opposite of putting on socks and shoes is taking the shoes off, followed by taking the socks off.

Now, a similar law holds for the transpose, namely:

$$(AB)^T=B^TA^T $$

for matrices $A,B$ such that the product $AB$ is defined. My question is: is there any intuitive reason as to why the order of the factors is reversed in this case?

[Note that I'm aware of several proofs of this equality, and a proof is not what I'm after]

Thank you!

Solution 1:

One of my best college math professor always said:

Make a drawing first.

Although, he couldn't have made this one on the blackboard.

Solution 2:

By dualizing $AB: V_1\stackrel{B}{\longrightarrow} V_2\stackrel{A}{\longrightarrow}V_3$, we have $(AB)^T: V_3^*\stackrel{A^T}{\longrightarrow}V_2^*\stackrel{B^T}{\longrightarrow}V_1^*$.

Edit: $V^*$ is the dual space $\text{Hom}(V, \mathbb{F})$, the vector space of linear transformations from $V$ to its ground field, and if $A: V_1\to V_2$ is a linear transformation, then $A^T: V_2^*\to V_1^*$ is its dual defined by $A^T(f)=f\circ A$. By abuse of notation, if $A$ is the matrix representation with respect to bases $\mathcal{B}_1$ of $V_1$ and $\mathcal{B}_2$ of $V_2$, then $A^T$ is the matrix representation of the dual map with respect to the dual bases $\mathcal{B}_1^*$ and $\mathcal{B}_2^*$.