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Good afternoon,

I need your help please in this question, and I'm sorry if my question is evident;

Let $X, Y\in M_n(\mathbb{R})$ be two real matrices and $I$ be the identity matrix. It is clear that $$(X-I)(Y-I)=XY-X-Y+I$$

The question is: show that if $XY=X+Y$, then $I-X$ is invertible?

What I have done: if $XY=X+Y$ then one has $(X-I)(Y-I)=I$, how I continue?

PS: * I know that a matrix $X\in M_n(\mathbb{R})$ is invertible if and only if there exist $X'\in M_n(\mathbb{R})$ such that $$XX'=X'X=I$$ ** Also we know that a matrix $X\in M_n(\mathbb{R})$ is invertible if and only if $|X|\neq0$.

Thank you

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    If you have understood that $(X−I)(Y−I)=I$ then isn't is obvious that $(I−X)(I-Y)=I$ and hence $(I−X)$ is invertible – 2017-01-17
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    my problem that is the multiplication between matrices is not commutative. Moreover I have well understood that $(I-X)(I-Y)=I$, but from this how we deduce that $(I-X)$ is invertible? as I said in PS a matrix is invertible iff $XX'=X'X=I$ – 2017-01-17
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    Just expand $(I−Y)(I−X)(I−Y)$ and substitute $XY=X+Y$. You will get the other equation. – 2017-01-17
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    @TushantMittal Thank you, now I got it. Im so sorry if my question was evident – 2017-01-17

2 Answers 2

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If $ XY = X+Y, $ then $$ (X-I)(Y-I) = XY - X - Y + I = X + Y - X - Y + I = I $$ So $$(X-I)^{-1} = (Y-I)$$ and clearly if $(X-I)$ is invertible so is $(I-X) = - (X - I)$. We also have to show that $(Y-I)$ is a left inverse of $(X-I)$, but this follows quite easily.

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    this is the objective of my question. M is invertible iff MM'=M'M=I; what you said in, you answer is MM'=I then M is invertible? Here I dont understand, you didnt proved that M'M=I – 2017-01-17
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If $PQ=I$ then $\det(PQ)=\det(I)$, and hence $\det(P)\cdot \det(Q) = 1$. It follows that $\det(P) \neq 0$ and $\det(Q) \neq 0$, i.e. both $P$ and $Q$ are invertible. Moreover, $P^{-1}=Q$ and $Q^{-1}=P$.

If $R$ is an $n$-by-$n$ matrix then $\det(-R)=(-1)^n\cdot \det(R)$ and hence $\det(R) \neq 0 \iff \det(-R) \neq 0$, i.e. $R$ is invertible if and only if $-R$ is invertible.

You have shown that $(X-I)(Y-I)=I$. If $P=X-I$ and $Q=Y-I$ then we can conclude that both $X-I$ and $Y-I$ are invertible. Moreover, it follows that both $I-X$ and $I-Y$ are invertible.

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    Non I said that MM'=M'M=I , this is my problem* – 2017-01-17
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    @As But it's the same. If $MM'=I$ then $M'M=I$. – 2017-01-17
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    this is my problem; I know that M is invertible iff MM'=M'M=I, my problem with the first equality. – 2017-01-17
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    the multiplication between matrices is not commutative – 2017-01-17
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    @As I know. But the inverse element of a group is unique, so if $MA=I$ and $BM=I$ then $B=A$. A matrix $X$ and its inverse $X^{-1}$ **do commute**, so $MM'=I \iff M'M = I$. The equation $MM'=M'M=I$ is not two conditions, it is only one. As I say above: if $PQ = I$ then $\det(PQ)=\det(I)$ and so $\det(P)\cdot \det(Q) = 1$. Hence both $P$ and $Q$ are both invertible. Moreover, $P^{-1} = Q$, $Q^{-1}=P$, and $QP=I$. – 2017-01-17
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    Thank you very much for your efforts , I got it; Thank you – 2017-01-18
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    @As I'm glad I could help. Please consider up-voting answers that you find "useful" by clicking the grey up-arrow, down-voting answers that were not useful by clicking the grey down-arrow, and selecting your favourite answer by clicking the "tick". – 2017-01-18