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If $A \in M_n(R) $ with $A^3=0$, show that $I - A$ is non-singular with $(I-A)^{-1}=I+A+A^2$.

How could I approach this?

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  • $\begingroup$ I tried to prove that (I-A)*(I+A+A^2) = I but my result was 0=I $\endgroup$ – Mark Ruiz Oct 2 '16 at 3:58
  • $\begingroup$ What does it mean that the product of $(I - A)$ with another matrix is the identity matrix? (What is the definition of (non-)singularity?) $\endgroup$ – TMM Oct 2 '16 at 4:03
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Notice that $(I-A)(I + A + A^2) = I + A + A^2 - A - A^2 - A^3 = I - A^3 = I$ since $A^3 = 0$

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  • $\begingroup$ Thank you, for some reason I mistakingly thought I^2=0 when my notes say otherwise. I see my mistake now. $\endgroup$ – Mark Ruiz Oct 2 '16 at 4:05
  • $\begingroup$ Of course. Notice that $I$ is functioning as $1$ here. $\endgroup$ – 3-in-441 Oct 2 '16 at 4:07
  • $\begingroup$ Thank you, it definitely looks a lot clearer now that I see my mistake. $\endgroup$ – Mark Ruiz Oct 2 '16 at 4:16
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For any martices $A,B$,

If $\ A B=I$, $A=B^{-1} \& B=A^{-1}$

$\ (I-A)(I+A+A^2)=I^3-A^3$ as $AI=IA=A$

Since $A^3=0$, $(I-A)(I+A+A^2)=I$

Thus, $(I-A)^{-1}=I+A+A^2$.

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