Matrices: left inverse is also right inverse? [duplicate]

Solution 1:

Since $AB=I$ then $B=B(AB)=(BA)B$. Note from $AB=I$ that $1=\det(AB)=\det(A)\det(B)$ so $\det(B)\neq0$.

So by $(BA)B=B$ we have:

$(BA-I)B=0$. Since $\det(B)\neq0$ then $B$ is not a $0$ divisor. So $BA=I$

Solution 2:

I suggest proving it in one line: Let $B\in\mathbb F^{n\times n}$ be right inverse, $C\in\mathbb F^{n\times n}$ left inverse of $A\in\mathbb F^{n\times n}$. Since Multiplying matrices is associative: $$B=IB=(CA)B=CAB=C(AB)=CI=C$$ Thus $B=C$ as required.

Solution 3:

This is true for linear transformations, and thus also for matrices.

EDIT: $AB=I\Rightarrow BAB=B\Rightarrow BABB^{-1}=BB^{-1}=I\Rightarrow BA=I$