Linear and multilinear algebra; matrix theory Proposition

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.

Reading path

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Identity matrix object
Matrix definition
Matrix multiplication definition
Square matrix definition
Invertible matrix definition
One-sided inverse of a square matrix is two-sided statement

Tag	`0753Y2X5K6`
URI	`m2://0753Y2X5K6`
Kind	statement, proposition
Writing	W0 stub
MSC	15A09
Status	This copy of the site has no build data. Run `m2 status 0753Y2X5K6`.
Uses	Invertible matrix

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

Statement

Remarks

Reading path

Further reading

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