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I have two square matrices $X$ and $Y$. Given :$X = I - XY$ and $Y^3 = 0$ how can I show that $X = I - Y + Y^2$ I tried raising to the power of $3$ both sides of the first equation but it didn't help much.

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    $\begingroup$ Write $X(I+Y)=I$, and $I=I-Y^3=\cdots$. $\endgroup$ – Dietrich Burde Nov 17 '17 at 17:04
  • $\begingroup$ Yes, X is invertible and X's inverse is (I + Y) but even if I raise this equation to the power of three it gets me nowhere $\endgroup$ – eventhorizon02 Nov 17 '17 at 17:06
  • $\begingroup$ so I wrote X(I + Y) = I - Y^3 but then I get to X = I - XY - Y^3 $\endgroup$ – eventhorizon02 Nov 17 '17 at 17:29
  • $\begingroup$ still not able to prove it, any other clue? $\endgroup$ – eventhorizon02 Nov 17 '17 at 18:07
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We have $(I+Y)(I-Y+Y^2)=I+Y^3=I$, so $(I+Y)^{-1}=I-Y+Y^2$. Now $X(I+Y)=I$, so $X=(I+Y)^{-1}$.

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  • $\begingroup$ Thank you very much. I understand we basically show that (I - Y + Y^2) is the inverse of (I + Y) and since X is also the inverse of (I + Y) and there is only one inverse we get that X = I - Y + Y^2 $\endgroup$ – eventhorizon02 Nov 17 '17 at 19:55
  • $\begingroup$ For the general formula for the inverse of $I+Y$ see here, or here. $\endgroup$ – Dietrich Burde Nov 17 '17 at 19:57
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$X=1-XY$ gives $XY^2=Y^2-XY^3=Y^2$

Also $X=1-XY$ gives $XY=Y-XY^2=Y-Y^2$

So $X=1-XY= 1-Y+Y^2$

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