Question: Let A be an m by n matrix with rank m . Prove that there exists an n b

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Question: LetA be an m by n matrix withrank m. Prove that there exists an n by mmatrix B such thatAB=Im. (Hint: Think about solvingAx=ek for k = 1, ... ,m.)

Solution my tutor and I came upwith:
from the given I can conclude that the rank(A) = m and rank(A) isless than equal to n.

If rank(A) = m = n
then A is invertible and so it is understood that B could equalA-1 so that AB = Im

If rank(A) = m < n

then for all k less than or equal to m ABk =ek
since rank(A) < n then
0 < n - rank(A)
so there must be a solution for Bk;therefore B exists so AB = I.

This made sense to me before. I thought that since the nullity ofthe matrix A is positive, then there must be at least one solutionfor B. But this doesn't make sense, because the dimension of A is mnot n.

Help!!! My exam istomorrow.

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