Why is the left inverse of a matrix equal to the right inverse? Given a square matrix $A$ that has full row rank we know that the matrix is invertible. So there is a matrix $B$ such that
$$
AB=1
$$
writing this in component notation,
$$
A_{ij}B_{jk}=\delta_{ik}
$$
Now, we tend to write $A^{-1}$ instead of $B$ but let's leave it like that for now.
My question is how can we show that $BA=1$? We mechanically jump to the conclusion that if the inverse exists, $AA^{-1}=A^{-1}A=1$ but how to show that? Equivalently why is the left inverse equal to the right inverse? It seems intuitively obvious!
Thanks a bunch, I appreciate.
 A: So you have a right inverse, and you know there is a left inverse too, let's say $C$. Then you have: $$1 = AB,$$ and multiplying both sides for C in the left, you get $$ C = C(AB) = (CA)B = B,$$ that is, if both exist, they must be equal.
A: Let the endomorphism
$$\Phi:\mathcal M_n(\Bbb R)\to\mathcal M_n(\Bbb R),\; X\mapsto XA$$
then $\Phi(X)=0\iff XA=0\iff XAB=X=0$ hence $\Phi$ is injective and by the rank-nullity theorem it's bijective hence surjective so there's $B'$ such that $\Phi(B')=I_n\iff B'A=I_n$ which means that $A$ has a left inverse. It's routine to prove that $B=B'$.
A: I just had a thought, using the commutator matrix $C=AB-BA=1-BA$. Post-multiply by $B$ we get
$$
CB=B-BAB=B-B=0
$$
using $AB=1$. Now $B$ has full row rank and therefore $C=0$ implying that $A$ and $B$ commute.
Is that a way to go or did I miss something?
EDIT:
Better yet, just consider the matrix $S=ABA$.
$$
S=(AB)A=A
$$
on the other hand
$$
S=A(BA)
$$
Therefore
$$
A=A(BA) \Rightarrow A(BA-1)=0 \Rightarrow BA=1
$$
A: Given $AB=I$ then $B=BI=B(AB)=(BA)B$  Since $B=(BA)B$ then $BA=I$.
A: Lemma: Suppose there is a $v_1\ne0$ so that $Av_1=0$, then the column space of $A$ has dimension at most $n-1$.
Proof: Suppose there is a $v_1\ne0$ so that $Av_1=0$. Create a basis $\{v_k\}_{k=1}^n$ for $\mathbb{R}^n$ including $v_1$. We can write all vectors in $\mathbb{R}^n$ as
$$
\sum_{k=1}^nc_kv_k
$$
which means we can write all vectors in the column space of $A$ as
$$
\sum_{k=2}^nc_kAv_k
$$
Therefore, the column space of $A$ has dimension at most $n-1$.
$$\square$$
Since $AB=I$, the column space of $A$ has dimension $n$. Therefore, $Ax=0\implies x=0$.
If $AB=I$, then $ABA=A$. Therefore, $A(BA-I)=0$. This means that each column of $BA-I$ is $0$. That is,
$$
BA=I
$$

Note that this depend on the vector space having finite dimension. Consider the two linear operators on sequences, $A$ and $B$, where $A$ shifts left:
$$
(Av)_k=v_{k+1}
$$
and $B$ shifts right, filling in with $0$:
$$
(Bv)_k=\left\{\begin{array}{}
0&\text{if }k=0\\
v_{k-1}&\text{if }k\ge1
\end{array}\right.
$$
We have $AB=I$, but $BA\ne I$ since $BA$ sets the first element of any sequence to $0$.
