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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.

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    $\begingroup$ Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible. $\endgroup$
    – coffeemath
    Apr 19, 2017 at 23:23
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    $\begingroup$ 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, 2017 at 23:26
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    $\begingroup$ @Coffeemath Why not same $\endgroup$ Apr 19, 2017 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, 2017 at 23:37
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    $\begingroup$ 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, 2017 at 23:38

2 Answers 2

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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.

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    $\begingroup$ 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, 2018 at 4:37
  • $\begingroup$ If yes, then please explain! $\endgroup$
    – Learning
    Jan 19, 2018 at 4:39
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    $\begingroup$ Yes, $X,Y$ are invertible matrices and these form a group. $\endgroup$
    – lulu
    Jan 19, 2018 at 10:47
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    $\begingroup$ @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, 2018 at 15:14
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    $\begingroup$ @TheStudent Sure, it's just notation. $\endgroup$
    – lulu
    Nov 24, 2019 at 14:14
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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

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    $\begingroup$ $XY=0$ 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, 2018 at 12:53
  • $\begingroup$ @obareey thanks, I have re-edited $\endgroup$
    – Revc_Ra
    Dec 26, 2018 at 15:59

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