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Are there two matrices $A_{m\times n}$, and $B_{n\times m}$ such that $A.B=I_{m}$ and $B.A=I_{n}$ (here $m\neq n$)

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@William: See also:… – InterestedGuest Mar 19 '11 at 22:11

No. This follows from the facts that $rk(AB) \leq \min(rk(A),rk(B))$, that $rk(I_n) = n$, and that for an $n \times m$ matrix $A$ we have $rk(A) \leq \min(m,n)$. Here $rk(\cdot)$ denotes the rank of a matrix.

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