# If $A+B = AB$ then proving that $AB = BA$?

Even though this has been asked before in main site If $A+B=AB$ then $AB=BA$, still I had a query?

If $A,B$ are both $n \times n$ matrix and the entries are from $\Bbb{R}$. If it satisfies $A+B = AB$, then can we say that $A$ and $B$ commute? that is $AB = BA$ ?

I thought of this that as we are given the rule $A+B = AB$, so that also implies that $B+A =BA$ just interchanging the roles of $A$ and $B$? from which we get $AB = BA$ hence $A$ and $B$ commute. Any flaw in this?

How do we approach this problem?

• it is not given that $X+Y=XY$ for all matrices, so that you can plug in $A$ and $B$ as you wish. It is only given that $A+B=AB$ for particular matrices $A$ and $B$. – Inactive - Objecting Extremism Aug 25 '18 at 12:51
• $\boldsymbol A, \boldsymbol B$ are fixed here, the symmetry might not be applicable here. – xbh Aug 25 '18 at 12:51
• $A+B=AB$ is not an identity so we can't change the roles of $A$ and $B$ and arrive at another correct equation. For example if you are solving $2x+3y=0$, then we can't interchange $x$ and $y$ and derive another equation $2y+3x=0$. – Marco Aug 25 '18 at 12:53
• I don't see where this question differs from the linked question. – Gerry Myerson Aug 25 '18 at 13:36
• @GerryMyerson The difference, I think, is that the OP is asking whether a certain specific proof is valid ($A+B=AB$ implies $BA=B+A=A+B=AB$.) – David C. Ullrich Aug 25 '18 at 14:45

The comments, especially Marco's, identify the fundamental flaw in the reasoning. You can interchange symmetrically equivalent variables only in an identity. Here the relevant identity is actually a logical expression:

$(A-I)(B-I)=I\implies (B-I)(A-I)=I$

That is where you can interchange the symetrically equivalent variables.

Incidentally, the original claim is not really true for all pairs of matrices satisfying the hypothesis. For finite order matrices, yes; but consider the infinite order matrices defined thusly, for all natural numbers $i, j$:

$A_{i,i}=A_{i,i+1}=1, A_{i,j}=0 \text{ otherwise}$

$B=A^t$

Then $AB=A+B$ but $BA\ne AB$. We get $(AB)_{1,1}=2$ but $(BA)_{1,1}=1$. Delving a little bit deeper, we find that $(A-I)(B-I)=I$ but $(B-I)(A-I)\ne I$ due to the latter having $0$ rather than $1$ in the $(1,1)$ position. The identity on which a proper proof is based, does not generally hold for infinite order matrices, and the claim goes with it.