So I can do a mathematical proof of $(AB)^{-1} = B^{-1}A^{-1}$. But I would like a deeper, I guess more applicable, understanding.
Here's my mathematical proof: $$(AB)^{-1}(AB) = I$$ $$(AB)^{-1}AB = I$$ $$(AB)^{-1}ABB^{-1} = IB^{-1} = B^{-1}$$ $$(AB)^{-1}AI =(AB)^{-1}A = IB^{-1} = B^{-1}$$ $$(AB)^{-1}AA^{-1} = B^{-1}A^{-1}$$ $$(AB)^{-1}= B^{-1}A^{-1}$$
I want to say that this intuitively makes sense because we need the rows and columns of both sides of the equation to equal. Let's say $A$ is a $n\ x\ k$ matrix and $B$ is a $k\ x\ n$, then $AB$ is a $n\ x\ n$ matrix and our $(AB)^{-1}$ matrix is also $n\ x\ n$.
Now, $B^{-1}$ is going to be a $n\ x\ k$ matrix while $A^{-1}$ is a $k\ x\ n$ matrix. It can't be that $(AB)^{-1} = A^{-1}B^{-1}$ because that's a $k\ x\ k$ matrix on the right.
However although this makes intuitive sense, this shouldn't hold up. B and A can't be non-square matrices because those can't have inverses. Only square matrices have inverses(non-zero determinant).
Can someone explain why $(AB)^{-1}= B^{-1}A^{-1}$ is true intuitively?
For me the fact that $B^{-1}A^{-1}$ "works" as a two-sided inverse to $AB$ is both intuitive, and somewhat rigorous.
Interestingly, as far as I know, the same identity, which holds in group theory more generally, is called the "socks and shoes identity".