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.
Asked
Active
Viewed 1.5k times
19
-
7Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible. – coffeemath Apr 19 '17 at 23:23
-
8The 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). – Apr 19 '17 at 23:26
-
1@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
-
6Just 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
2 Answers
72
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.
lulu
- 70,402
-
1if $ 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 04:37
-
-
1
-
but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$ – Learning Jan 19 '18 at 15:10
-
2@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
-
-
1
-
@lulu, do we have any requirement for =(−)(−) to hold? OP's question doesn't make any assumptions on the field or group of and objects. I cannot understand why (−)(−)= would imply (−)(−) =. Basically, I'm missing why "Thus (−) and (−) are inverse to each other, whence". Could you elaborate on that a little? – Alt Sep 06 '22 at 14:43
-
In a group, left inverses coincide with right inverses, see, e.g., this Invertible matrices form a group. I am assuming that these are matrices over a field, haven't thought about generalizations (which presumably exist). – lulu Sep 06 '22 at 14:49
3
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
-
3$XY=0$ 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
-