If AB=I then BA=I

Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.
Suppose each matrix is n by n. We consider our matrices to all be acting on some n-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and n by n matrices).
Then AB has range equal to the full space, since AB=I. Thus the range of B must also have dimension n. For if it did not, then a set of n1 vectors would span the range of B, so the range of AB, which is the image under A of the range of B, would also be spanned by a set of n1 vectors, hence would have dimension less than n.
Now note that B=BI=B(AB)=(BA)B. By the distributive law, (IBA)B=0. Thus, since B has full range, the matrix IBA gives 0 on all vectors. But this means that it must be the 0 matrix, so I=BA.

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