If $A^n=0$, then $I_n-A$ is invertible. How do I solve this problem?
$A$ is $n\times n$ and $A^n=0$. Prove that  $I_n-A$ is invertible.
 A: Consider the matrix $B := I_n + A + A^2 + \cdots + A^{n-1}$. Show that $(I_n - A)B = I_n$ using the relation $A^n = 0$.
A: $$(I-A)(I+A+A^2 + \dots + A^{n-1}) =$$$$= I+A+A^2 + \dots + A^{n-1} - A-A^2 - \dots - A^n=$$
$$= \mbox{ everything cancels out} =I-A^n = I$$
So $I-A$ is invertible.
A: If $I_n-A$ is not invertible, then it is not a $1-1$ transformation $\mathbb R^n\to\mathbb R^n$, so it must have a non-zero vector $v$ such that $(I_n-A)v=0$. Then $Av=v$ and $A^nv=v$ by induction. Since $A^n=0$, this means $v=0$, contradicting the above assumption.
The algebraic proofs have the advantage that they necessarily work on infinite-dimension vector spaces. If $T:V\to V$ is a linear transformation and $T^k=0$ for some $k$, then the proof above shows that $I_V-T$ is invertible. My proof does not, because a map $V\to V$ can be $1-1$ and not be invertible when $V$ is infinite-dimensional.
A: These algebraic proofs can be extended to a stronger claim:

The spectrum of a nilpotent operator contains only the identity (c.f. this). 

Reproducing Geoff Robinson's fine answer to that question, which mirrors answers above, for any complex $\lambda\neq 0$, 
$$\lambda^n I = \lambda^n I - A^n = (\lambda I - A)(\lambda^{n-1}I + \lambda^{n-2}A + \lambda^{n-3}A^2 + \cdots + A^{n-1}) $$
so this constructs an inverse to the resolvent $\lambda I - A$. 
In particular, $\lambda = 1$ answers this question.
A: Here is another argument. Since $A^n=0$, all eigenvalues of $A$ are zero. So the Jordan form $J$ of $A=SJS^{-1}$ has all zeroes in its diagonal. Then $I-J$ is upper triangular with all the diagonal elements nonzero, so invertible. Thus
$$
I-A=S(I-J)S^{-1}
$$
is invertible. 
