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
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    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
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    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

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