# If $A+B=AB$ then $AB=BA$

I was doing the problem $$A+B=AB\implies AB=BA.$$

$AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.

• Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible. – coffeemath Apr 19 '17 at 23:23
• The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^{-1}$ exists). – user228113 Apr 19 '17 at 23:26
• @Coffeemath Why not same – Sachchidanand Prasad Apr 19 '17 at 23:29
• If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA. – coffeemath Apr 19 '17 at 23:37
• Just to say, suppose $A$ and $B$ are both the zero matrix. Then of course $A+B=AB=BA$ but neither $A$ nor $B$ is invertible. – lulu Apr 19 '17 at 23:38

Consider the expression $$(A-\mathbb 1)(B-\mathbb 1)=AB-A-B+\mathbb 1=\mathbb 1$$

Thus $(A-\mathbb 1)$ and $(B-\mathbb 1)$ are inverse to each other, whence $$\mathbb 1= (B-\mathbb 1)(A-\mathbb 1)=BA - A - B + \mathbb 1$$

It follows that $$BA=A+B=AB$$ and we are done.

Note: here $\mathbb 1$ denotes the appropriate identity matrix.

• if $a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ? – Learning Jan 19 '18 at 4:37
• If yes, then please explain! – Learning Jan 19 '18 at 4:39
• Yes, $X,Y$ are invertible matrices and these form a group. – lulu Jan 19 '18 at 10:47
• @Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=\mathbb 1$. If nothing else, that equation shows that $\det X \times \det Y =1\neq 0$ so both $X,Y$ have to be invertible, – lulu Jan 19 '18 at 15:14
• @TheStudent Sure, it's just notation. – lulu Nov 24 '19 at 14:14

Start with equation: $$A+B=AB$$ replace $$B$$ with $$(B - I) + I$$ in left side $$A+(B - I) + I =AB$$ $$I =AB - A - (B - I)$$ $$I =A(B - I) - (B - I)$$ $$I =(A - I)(B - I)$$ the inverse matrix of $$(B - I)$$ is $$(A - I)$$, so $$(B - I)$$ is invertible. Then the rest follows as in previous answer by @lulu

• $XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $Y=\begin{bmatrix}0&0\\0&1\end{bmatrix}$ – obareey Dec 26 '18 at 12:53
• @obareey thanks, I have re-edited – Revc_Ra Dec 26 '18 at 15:59