Let $A$ and $B$ be two $n$-by-$n$ real matrices such that $A+B = AB$. How do I prove that $AB= BA$?

I have tried using the trace function on $A+B-AB$. But I could not get any Ideas. Kindly provide me with hints.

  • 2
    I want to remark that, for a vector space $V$ over a field $K$ and for two $K$-linear operators $A,B: V\to V$ such that $A+B=A\circ B$, if $V$ is finite-dimensional over $K$, then $A$ and $B$ commute (i.e., $A\circ B=B\circ A$). However, the result does not hold if $V$ is infinite-dimensional over $K$. – Batominovski Oct 12 at 16:44
  • 2
    Take, for example, $V$ to be the vector space of infinite sequences $\mathbf{x}:=\left(x_i\right)_{i\in\mathbb{Z}_{>0}}$ of elements $x_1,x_2,x_3,\ldots\in K$. For each $\mathbf{x}:=\left(x_i\right)_{i\in\mathbb{Z}_{>0}}\in V$, define $$A(\mathbf{x}):=\left(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_5,x_5-x_6,\ldots\right)$$ and $$B(\mathbf{x}):=\left(x_1,x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,\ldots\right)\,.$$ – Batominovski Oct 12 at 16:45
  • 2
    Then, for every $\mathbf{x}:=(x_i)_{i\in\mathbb{Z}_{>0}}$ in $V$, $$(A+B)(\mathbf{x})=\left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,\ldots\right)$$ and $$(A\circ B)(\mathbf{x})=A\left(x_1,x_2-x_1,x_3-x_2,\ldots\right)=\left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,\ldots\right)\,,$$ but $$(B\circ A)(\mathbf{x})=B\left(x_1-x_2,x_2-x_3,x_3-x_4,\ldots\right)=\left(x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,\ldots\right)\,.$$ Thus, $$A+B=A\circ B\text{ but }A\circ B\neq B\circ A\,.$$ – Batominovski Oct 12 at 16:45
  • Thanks for your insight. It is very useful. – tony Oct 12 at 17:31
up vote 12 down vote accepted

$A+B=AB$ is equivalent to $(I-A)(I-B)=I$. As $I-A$ and $I-B$ are square, this implies $(I-B)(I-A)=I$, etc.

  • @tony then mark it as answered. – user25959 Oct 12 at 16:36
  • how do i do that? – tony Oct 12 at 16:38
  • 1
    @user25959 You cannot accept an answer before 15 min. after the post. – cansomeonehelpmeout Oct 12 at 16:38

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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