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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.

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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.

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    $\begingroup$ "The" is misleading. There are lots of such $B$'s to be obtained in this way. $\endgroup$ Jan 14 at 18:15

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