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 February 24th, 2010, 11:32 PM #1 Senior Member   Joined: Nov 2009 Posts: 169 Thanks: 0 Prove there is a matrix X Let A be an mxn matrix and suppose that B is an nxp matrix such that every column of B is in the column space of A. Prove that there is a matrix X such that AX=B. (HITN: we know the equation Ax=v has a solution if and only if v $\in$ col(A))
 February 25th, 2010, 05:21 AM #2 Newbie   Joined: Jan 2010 Posts: 21 Thanks: 0 Re: Prove there is a matrix X It doesn't make sense. B should be MxP. Anyway, assuming that B is MxP matrix : {v1,v2,...,vm} in Rn are the columns of A. From the question, if b1 is a column of B then b1 = a1v1+a2v2+...+amvm, where a1,a2,...,am in F. In other words, x1 = [a1,a2,...,am] is a solution to this system : Ax1=b1. That brings us to Axi=bi, where 1<=i<=p. Therefore, if you'll take all the x1,...xp and put them as columns in a matrix you'll get X such that AX=B.

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