Prove that the order of multiplication for two square matrices whose product is equal to the identity matrix does not matter The following proof is supposed to show that the order of multiplication for two square matrices whose product is equal to the identity matrix does not matter. However, although I understand the calculations of the proof, I cannot see at all how it confirms that the previous statement is true.



I would greatly appreciate it if someone could clarify for me how it exactly proves this statement. Also, please use lower-level mathematical lingo, as I am (obviously) not advanced when it comes to mathematics. :) Thank you.
 A: I hope this will clarify it. 
Let $AB=I.$ 
(1).   $Bv=Bw\implies v=w.$ Because$ Bv=Bw\implies B(v-w)=0\implies 0=A(B(v-w))=(AB)(v-w)=I(v-w)=v-w.$
(2).  Let $\{v_1,...,v_n\}$ be a linearly independent set of vectors. Then $S=\{Bv_1,...,Bv_n\}$ is a linearly independent set of vectors. Because if $a_1,...,a_n$ are scalars with not all of them $0,$ then $0\ne\sum_{j=1}^na_jv_j $ implies (by (1) ) that $B(0)\ne B(\sum_{j=1}^na_jv_j).$ That is, $0=B(0)\ne \sum_{j=1}^n a_jBv_j.$ 
(3). Therefore $S$ is a vector-space basis, so every vector $v$ is of the form $\sum_{j=1}^na_jBv_j=B(\sum_{j=1}^na_jv_j)=B(x_v).$ 
(4). Finally,since for any vector $v$  there exists $x_v$ such that $v=Bx_v,$ we have, for every $v$,  $$(BA-I)v=(BA-I)(Bx_v)=(BAB-B)x_v=(B(AB-I))x_v=$$ $$=(B\cdot O)x_v=(O)\cdot x_v=0.$$ So $(BA-I)x_v=0$ for all $v,$ so $BA-I=O.$
(5). We can also do step (4) as follows: Suppose, by contradiction,that $v\ne BAv$ for some $v.$ Then $Bx_v\ne BA(Bx_v)=B(ABx_v)=B(Ix_v)=Bx_v,$ which is absurd.
Remark. It is necessary to use the fact that we have a finite-dimensional vector space. In an infinite-dimensional vector space $V$ there are linear functions $A:V\to V$ and $B:V\to V$ with $ABv=v$ for all $v\in V$ but $BAv\ne v$ for some $v\in V.$ 
A: What the second part says is that if $f\circ g=0$ and $g$ is surjective, then $f$ must be zero. The reason: since $g$ is surjective, every member $v$ in the domain of $f$ is an image of some $x$ under $g$; hence $f(v)=f(g(x))=(f\circ g)(x)=0$. In particular, we have $g(x)=Bx$ and $f(v)=(BA-I)v$ in your proof.
If you read the proof carefully, you will see that $g$ is surjective because it is injective (in the first paragraph of the proof, it is first shown that $Bv=0\Rightarrow ABv=0\Rightarrow v=Iv=0$, i.e. $g$ is injective; then the rank-nullity theorem is applied to show that the column space of $B$ is the whole $\mathbb R^n$, i.e. $g$ is surjective). This "invectiveness implies subjectiveness" for linear maps actually relies on finite dimensionality of the vector spaces in question. For infinite-dimensional vector spaces, the implication $AB=I\Rightarrow BA=I$ no longer holds. See the popular thread "If $AB = I$ then $BA = I$" for an in-depth discussion.
