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 isn 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).

ThenAB 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 n−1 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 n−1 vectors, hence would have dimension less than n .

Now note thatB=BI=B(AB)=(BA)B . By the distributive law, (I−BA)B=0 . Thus, since B has full range, the matrix I−BA gives 0 on all vectors. But this means that it must be the 0 matrix, so I=BA .

Suppose each matrix is

Then

Now note that

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