If AB=IAB = I then BA=IBA = I

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.

  1. Elementary row operations.
  2. Linear dependence.
  3. 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.


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.

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

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