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I know how to prove from $AB=A^2+A+I_n$ (with $A,B\in \mathcal{M}_n(\mathbb{R})$) that $AB=BA$ by showing that A is invertible. I would like to know if one can derive $AB=BA$ without using that $A$ is invertible.

user37238
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  • Can you show that $A^2 \times A = A \times A^2$? – Calvin Lin Dec 14 '23 at 00:45
  • I suppose that if the invertibility of $A$ isn't allowed neither is using that $nullity(A)=0$, right? – Deif Dec 14 '23 at 01:24
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    $$\begin{aligned} AB\left(I-A\right)& =(I+A+A^{2})(I-A) \ AB-ABA& =I-A^{3} \ AB+A^{3}-ABA& =I \ AB+A\left(A-B\right)A& =I \ A(B+(A-B)A)&=I \ \end{aligned}$$ it implies that $|A|\neq0$ – Zhiwei Dec 14 '23 at 03:03
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    Let $X=A$ and $Y=B-A-I$. You are essentially asking how to prove that $XY=I\implies YX=I$ without taking matrix inverse. – user1551 Dec 14 '23 at 04:52
  • @CalvinLin Since the matrix product is associativ, yes. But how does this help? – user37238 Dec 14 '23 at 07:35
  • @Zhiwei Does $|A|\neq 0$ means $\det(A)\neq 0$? If so, you're using the invertibility of A... – user37238 Dec 14 '23 at 07:36
  • @user1551 My question is : on this particular exemple, with this particular identity, is there another way? – user37238 Dec 14 '23 at 07:36
  • @Deif Indeed, I'm looking for a another type of proof. – user37238 Dec 14 '23 at 07:58
  • @user37238 Yes, but I mean its a conclusion of $AB=A^2+A+I_n$,Because you take the determinant for both sides of $A(B+(A-B)A)=I$, you get that $\det(A)\neq 0$. – Zhiwei Dec 14 '23 at 08:07
  • @user37238 I don’t understand your response. To prove $XY=I\implies YX=I$ for square matrices $X$ and $Y$, there is no need to take matrix inverse. (In fact, the whole point of proving this assertion is to justify the existence of matrix inverse.) See the answers in the linked question. – user1551 Dec 14 '23 at 16:38

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