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
The essence of your proof is correct, but the first few lines are very confusing. Why not something more clear like:
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.
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 $BA$.