If A and B are square matrices such that AB=I, where I is the identity matrix, show that BA=I.

I do not understand anything more than the following.

- Elementary row operations.
- Linear dependence.
- Row reduced forms and their relations with the original matrix.
If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.

**Answer**

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 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 that B=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.

**Attribution***Source : Link , Question Author : Dilawar , Answer Author : Community*