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 $m\neq n$)
Product of Matrices
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linear-algebra
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0@William: See also: http://math.stackexchange.com/questions/17908/why-is-it-that-if-a-is-an-n-by-m-matrix-and-both-ba-and-ab-are-indentity-matrice – 2011-03-19
1 Answers
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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.