Let A be an m x n matrix with rank m. Prove that there exists an n x m matrix B

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Let A be an m x n matrix with rank m. Prove that there exists an n x m matrix B such that AB = In.

Explanation / Answer

Hello, A is an m x n matrix with rank m, rank is the column space of a matrix therefore we can say that n>=m because it would have to have m or greater columns in order to have m column space. Now notice that an m x n matrix times a n x m matrix will result in a m x m matrix. This matrix is square since it has the same amount of rows and columns, which is the identity matrix with a main diagonal of m 1's. Now to prove the n x m matrix exists. Take any n x m matrix B with rank n. rank AB = rank A = m, hence AB is invertible. Let M be the inverse of AB, then (AB)M = A(BM) = I, i.e. BM is the matrix n x m which was what we were trying to prove.

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Let A be an m x n matrix with rank m. Prove that there exists an n x m matrix B

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