Prove that A - I is nonsingular If matrix $A^3 = O$, then $A - I$ is nonsingular? True or false?
I tried to solve it like this: 
Given a linear transformation $T: V \rightarrow V$, such that $$m(T) = A, m(O) = O, m(I) = I, m(T^3) = A^3$$
where $m(.)$ is an operator that converts a linear transformation into the matrix. Assume that $(T - I)(x) = O$. Further assume that $x \neq O$. 
Then, $$(T - I)(x) = O \; \Rightarrow \; T(x) - I(x) = T(x) - x = O \;  \Rightarrow \; T(x) = x$$ for some $x \neq O$. So, $T(x) = I(x)$ for some $x$.
But we also have $$m(T^3) = A^3 = O \; \Rightarrow \; T^3(x) = O \; \forall x \in V$$ so $T^3(x) = I^3(x) = O$ which is a contradiction. Thus, $x = O$. 
Since $x = O$, the transformation $T - I$ is one-to-one ($(T - I)(x) = O$ implies $x = O$). 
Thus, $m(T - I) = A - I$ is nonsingular. 
Can someone check this proof? I found a solution online, which says this is an incorrect result. 
I would also be interested to see other proofs but you are not allowed to used determinants (Apostol chapter 2 calculus vol.2 is the limiter)! You can freely use the isomorphism of the space of linear transformations and matrices with respect to addition, scalar multiplication, and composition.

Hint:
If $A^3=0$, what is $(A-I)(A^2+A+I)$?
