If AB=I, then BA=I

Proposition: Let A and B be nxn matrices. Also, let

AB=I

then

BA=I

Proof: Every matrix is row equivalent to a single row-reduced echelon matrix. That is, for every matrix, there is a finite sequence of elementary row operations such that if we apply this sequence of row operations on this matrix, we get a row-reduced echelon matrix. If we denote R to be this matrix,

e_r(e_r-1(………e₁(R)…..))=A

for some elementary row operations, e_r,e_r-1,…….,e₁. Every elementary row operation on any matrix corresponds to multiplication of an elementary matrix with that matrix. This elementary matrix is obtained by doing the row operation on I_mxm where m is equal to row number of the matrix that we apply the elementary row operation. Denote E_i to be correspondent elementary matrix of e_i. Then,

E_rE_r-1…….E₁R=A

(E_rE_r-1…….E₁R)B=I

multiply both sides E_r^‘, inverse of E_r

E_r^‘(E_rE_r-1…….E₁R)B=E_r^‘I

keep multiplying,

E_r-1^‘E_r^‘(E_rE_r-1…….E₁R)B=E_r-1^‘E_r^‘I

E₁^‘……..E_r-1^‘E_r^‘(E_rE_r-1…….E₁R)B=E₁^‘……….E_r-1^‘E_r^‘I

RB=E₁^‘……….E_r-1^‘E_r^‘I

Can R contain a zero row? No, because if it were, then RB would contain a zero row and any matrix that is row equivalent to I can’t contain a zero row. Why? Because if there were such matrix, we would obtain a contradiction. Let’s assume that RB contains some zero rows and find (RB)’ (equal to RB without zero rows) matrix’s row-reduced echelon matrix, and then add zero rows at the end of this row-reduced echelon matrix. Then, this final matrix, say M, is row reduced echelon matrix of RB, and clearly it is not equal to I. This is a contradiction, since we have two distinct row reduced echelon matrix of RB; one is I and the other is M. This is not possible.

So, R must not have zero row, then it must be I (easy to verify). Then,

RB=IB=B=E₁^‘……….E_r-1^‘E_r^‘I

Is BA=I?

BA=(E₁^‘……….E_r-1^‘E_r^‘I)(E_rE_r-1…….E₁R)

=(E₁^‘……….E_r-1^‘E_r^‘I)(E_rE_r-1…….E₁I)=I

Yes, it is