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.

Your proof is correct, but there is no need to work so hard. (Note also that there is no “operator that converts a linear operator to a matrix” in the abstract; you have to specify an ordered basis, and you get the coordinate matrix relative to that basis).
You can operate direction with the matrix $A$, multiplying $n\times 1$ vectors. You know that $A^3\mathbf{x}=\mathbf{0}$ for all $\mathbf{x}$. So if $(A-I)\mathbf{x}=\mathbf{0}$, then as you note you have $A\mathbf{x}=\mathbf{x}$, so $A^2\mathbf{x}=A\mathbf{x}=\mathbf{x}$, and $A^3\mathbf{x}=A\mathbf{x}=\mathbf{x}$. Since $A^3\mathbf{x}=\mathbf{0}$, that means $\mathbf{x}=\mathbf{0}$. This proves $A-I$ is nonsingular: we have shown that if $(A-I)\mathbf{x}=\mathbf{0}$, then $\mathbf{x}=\mathbf{0}$.
