One-sided inverse of a square matrix is two-sided

Statement

Let $F$ be a field and let $A$ and $B$ be $n \times n$ matrices over $F$. If $BA = I_n$, then $AB = I_n$, so $A$ is invertible and $A^{-1} = B$. The same holds with the roles of $A$ and $B$ exchanged.

Remarks

The statement also holds over each commutative ring, by the determinant: $BA = I_n$ gives $\det B \det A = 1$. For non-square matrices it fails: a $2 \times 1$ matrix can have a left inverse but never a right inverse. A proof over a field uses rank–nullity: $BA = I_n$ makes $x \mapsto Ax$ injective, hence surjective.

Explanations

L1 human

For square matrices, we only need to check one of $AB = I$ and $BA = I$. The other one then holds automatically. This saves half of the work when we check an inverse.