# Showing that $I-A$ is non-singular

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?

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

Notice that $(I-A)(I + A + A^2) = I + A + A^2 - A - A^2 - A^3 = I - A^3 = I$ since $A^3 = 0$

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

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