How to prove that if $A$ and $B$ are two $n×n$ matrices with non negative integer entries such that $AB=I$.Then how to show $A$ and $B$ are permutation matrices.Permutation matrices are matrices with the columns just a permutation of the identity matrix. How do i proceed.Can somebody hint.
Matrices that can only be permutation matrices
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linear-algebra
matrices
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0See several proofs [here](http://mathoverflow.net/questions/62125/invertible-matrices-of-natural-numbers-are-permutations-why). – 2017-02-11
1 Answers
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Hint : Which pairs of vectors with non-negative integers have scalar-product $0$ or $1$ ?
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0Additional hint : Use $A^TB^T=(BA)^T=I^T=I$ and the row-sum-norm – 2017-02-11
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1i can see that if any of the entries in $A$ is greater than $1$ then since the product with entries from B has to result in a $0$ or $1$ then that column of B has to be zero which contradicts the invertibility of B – 2017-02-11
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0Excellent. Now rule out that we have more than one $1$ in any column or row. Finally rule out a zero-row/zero-column (almost trivial) – 2017-02-11