13
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible. $\endgroup$ – coffeemath Apr 19 '17 at 23:23
  • 5
    $\begingroup$ 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). $\endgroup$ – user228113 Apr 19 '17 at 23:26
  • $\begingroup$ @Coffeemath Why not same $\endgroup$ – Sachchidanand Prasad Apr 19 '17 at 23:29
  • $\begingroup$ If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA. $\endgroup$ – coffeemath Apr 19 '17 at 23:37
  • 3
    $\begingroup$ 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. $\endgroup$ – lulu Apr 19 '17 at 23:38
55
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $1$ ? $\endgroup$ – Learning Jan 19 '18 at 4:37
  • $\begingroup$ If yes, then please explain! $\endgroup$ – Learning Jan 19 '18 at 4:39
  • 1
    $\begingroup$ Yes, $X,Y$ are invertible matrices and these form a group. $\endgroup$ – lulu Jan 19 '18 at 10:47
  • 2
    $\begingroup$ @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, $\endgroup$ – lulu Jan 19 '18 at 15:14
  • 1
    $\begingroup$ @TheStudent Sure, it's just notation. $\endgroup$ – lulu Nov 24 '19 at 14:14
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ $XY=0$ 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 $Y=\begin{bmatrix}0&0\\0&1\end{bmatrix}$ $\endgroup$ – obareey Dec 26 '18 at 12:53
  • $\begingroup$ @obareey thanks, I have re-edited $\endgroup$ – Revc_Ra Dec 26 '18 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.