# Rows of A are linearly independent. Then there exists B such that AB = I

Suppose rows of matrix $$A_{m\times n}$$ are linearly independent. Prove that there exists matrix $$B_{n\times m}$$ such that $$AB = I_{m\times m}$$.

Well, basically it's asked to prove that there exists a right inverse of $$A$$, but I don't understand how'd I do that. I tried to prove it using elementary transformation matrices, but had no success.

Rows are independent $$\implies$$ $$m\le n$$, and you can "complete" the matrix $$A$$ to a $$n\times n$$ matrix $$A'$$ with maximum rank and so invertible. There exists the inverse of $$A'$$, denote it by $$B'$$ (always $$n\times n$$), $$A'B'=I_{n\times n}$$. Take $$B$$ the $$n\times m$$ matrix composed of the first $$m$$ columns of $$B'$$.
This is the situation: $$A'= \begin{pmatrix} A\\ * \end{pmatrix},\quad B'= \begin{pmatrix} B& * \end{pmatrix}, \quad A'B'= \begin{pmatrix} AB&*\\ *&* \end{pmatrix} =I_{n\times n}= \begin{pmatrix} I_{m \times m}&0\\ 0&I_{n-m\times n-m} \end{pmatrix}.$$
The matrix $$B$$ is the matrix you are looking for.
• "The" is misleading. There are lots of such $B$'s to be obtained in this way. Jan 14 at 18:15