0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Tushant Mittal Jan 17 '17 at 17:00
  • $\begingroup$ 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 $XX'=X'X=I$ $\endgroup$ – A s Jan 17 '17 at 17:07
  • $\begingroup$ Just expand $(I−Y)(I−X)(I−Y)$ and substitute $XY=X+Y$. You will get the other equation. $\endgroup$ – Tushant Mittal Jan 17 '17 at 17:16
  • $\begingroup$ @TushantMittal Thank you, now I got it. Im so sorry if my question was evident $\endgroup$ – A s Jan 17 '17 at 17:20
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – A s Jan 17 '17 at 17:17
-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Non I said that MM'=M'M=I , this is my problem* $\endgroup$ – A s Jan 17 '17 at 17:02
  • $\begingroup$ @As But it's the same. If $MM'=I$ then $M'M=I$. $\endgroup$ – Fly by Night Jan 17 '17 at 17:05
  • $\begingroup$ this is my problem; I know that M is invertible iff MM'=M'M=I, my problem with the first equality. $\endgroup$ – A s Jan 17 '17 at 17:14
  • $\begingroup$ the multiplication between matrices is not commutative $\endgroup$ – A s Jan 17 '17 at 17:14
  • 1
    $\begingroup$ @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". $\endgroup$ – Fly by Night Jan 18 '17 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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