Let $T : V \rightarrow V$ be a linear operator. If $T^n = O_V$ for some $n \ge 1$, prove that $I_V + T$ is an isomorphism. I think I'm supposed to prove that both $T$ and $I_V$ are isomorphic, but I'm not sure and even if that's the right way to go, how am I supposed to prove that $I_V$ is isomorphic. Or am I going about this all wrong. Please help.
 A: Hint: $(1+x)\left(1-x+x^2-\ldots+ (-x)^{n-1}\right) = 1-(-x)^n$.
A: A map $F: V \to V$ is an isomorphism by definition if there is $G: V \to V$ such that $FG$ and $GF$ are both identity maps. Now consider $(I_V + T)(I_V - T + T^2 - T^3 + \ldots + (-1)^{n-1} T^{n-1})$.
A: Hint: A endomorphism on a vector space is an isomorphism $\iff$ it is invertible $\iff$ it is surjective. So, can you prove that if $(I_V + T)v = 0$ then $v=0$?
Further hint: Let $m\ge 0$ be the least number such that $T^m v = 0$. This exists by the given property. Prove that $m=0$.
A: Since $T^n = 0$, it is easy to write down the inverse of $I_V + T$; indeed, we have
$(I_V + T)^{-1} = \sum_0^{n-1} (-1)^i T^i; \tag{1}$
here we understand that $T^0 = I_V$ and the sum is truncated after $i = n -1$ since $T^n = 0$.  To check:
$(I_V + T) \sum_0^{n -1}(-1)^i T^i = \sum_0^{n-1}(-1)^i T^i  + \sum_0 ^{n - 1} (-1)^i T^{i + 1}$
$= I_V + \sum_1^{n -1} ((-1)^i + ( -1)^{i -1}) T^i = I_V, \tag{2}$
where we have used the hypothesis $T^n = 0$ to drop the final $T^n$ from the summation on the right.  This shows $I_V + T$ is invertable. QED.
Hope this helps.  Cheers,  
and as always,
Fiat Lux!!!
