# Prove that if $A$ is an $n \times n$ matrix such that $A^{4} = 0$ then $(I_n - A)^{-1}=I_n+A+A^2+A^3$

Prove that if $$A$$ is an $$n \times n$$ matrix such that $$A^{4}$$ = 0 then: $$(I_n - A)^{-1}=I_n+A+A^2+A^3$$

My proof is as follows: $$(I_n - A)(I_n - A)^{-1}=I_n$$ $$(I_n - A)^{-1}=I_n/(I_n - A)$$ $$I_n/(I_n - A)=I_n+A+A^2+A^3$$ $$I_n=(I_n - A)(I_n+A+A^2+A^3)$$ $$I_n=I_n+A+A^2+A^3-A-A^2-A^3-A^4$$ $$I_n=I_n-A^4$$ because we know that: $$A^4=0$$ therefore: $$I_n=I_n$$

Is this an acceptable justification or have I made an error in my logic?

*I apologize for any poor formatting

• First of all, you shouldn't ever 'divide' by matrices. Only multiply with their inverse – Lukas Rollier Oct 5 '20 at 14:41
• @LukasRollier Right, thank you, I will go back and work through it again. – Nish Oct 5 '20 at 14:43

In order to show $$(I_n-A)^{-1}=I_n + A +A^2 +A^3$$, it suffices to show $$(I_n-A)(I_n + A +A^2 +A^3)=(I_n + A +A^2 +A^3)(I_n-A)=I_n.$$ It's easy to show that the left equals the middle. Then by multiplying through and using the fact that $$A^4=0$$, it's easy to show that both of them equal the right.
Hint: Let $$B=I+A+A^2+A^3$$. Compute $$AB$$.
To prove $$a$$ is inverse of $$b$$. Just find $$ab$$ and $$ba$$. If $$b$$ is inverse of $$a$$ then $$ab=I=ba$$. Thus Take $$B=I_n+A+A^2+A^3$$ and solve $$AB$$ and $$BA$$.