If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible? If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible?
This question yielded two different proofs from my professors, which managed to get conflicting results (true and false). Could you please weigh in and explain what's happening, and offer a working proof?
Proof that it is invertible: Consider matrix $A-I_{n}$. Multiplying $(A+I_{n})$ by $(A-I_{n})$
we get $A^2-AI_{n}+AI_{n}-I^2_{n}$. This simplifies to $A^2-I^2_{n}$ which is equal to $-I_{n}$, since $A^2=0$. So, the professor argued, since we have shown that there exists a $B$ such that $(A+I_{n})$ times $B$ is equal to $I$, $(A+I_{n})$ must be invertible. I am afraid, though, that she forgot about the negative sign that was leftover in front of the $I$ -- from what I understand, $(A+I_{n})$*$(A-I_{n})$=$-I$ does not mean that $(A+I_{n})$ is invertible.
Proof that it is not invertible: Assume that $A(x)=0$ has a non-trivial solution. Now, given $(A+I_{n})(x)=\vec{0}$, multiply both sides by $A$. We get $A(A+I_{n})(x)=A(\vec{0})$, which can be written as $(A^2+A)(x)=\vec{0}$, which simplifies to $A(x)=0$, as $A^2=0$. Since we assumed that $A(x)=0$ has a non-trivial solution, we just demonstrated that $(A+I_{n})$ has a non-trivial solution, too. Hence, it is not invertible.
I am not sure if I reproduced the second proof in complete accuracy (I think I did), but the idea was to show that if $A(x)=\vec{0}$ has a non-trivial solution, $A(A+I_{n})$ does too, rendering $A(A+I_{n})$ non-invertible. But regardless of the proofs, I can think of examples that show that at least in some cases, the statement is true; consider matrices $\begin{bmatrix}
0 & 0\\ 
0 & 0
\end{bmatrix}$ and $\begin{bmatrix}
0 & 1\\ 
0 & 0
\end{bmatrix}$ which, when added $I_{2}$ to, become invertible.
Thanks a lot!

I suggest thinking of the problem in terms of eigenvalues.  Try proving the following:
If $A$ is an $n \times n$ matrix (over any field) which is nilpotent -- i.e., $A^k = 0$ for some positive integer $k$, then $-1$ is not an eigenvalue of $A$ (or equivalently, $1$ is not an eigenvalue of $-A$).  
If you can prove this, you can prove a stronger statement and collect bonus points from Arturo Magidin.
(Added: Adrian's answer -- which appeared while I was writing mine -- is similar, and probably better: simpler and more general.  But I claim it is always a good idea to keep eigenvalues in mind when thinking about matrices!)
Added: here's a hint for a solution that has nothing to do with eigenvalues (or, as Adrian rightly points out, really nothing to do with matrices either.)  Recall the formula for the sum of an infinite geometric series:
$\frac{1}{1-x} = 1 + x + x^2 + \ldots + x^n + \ldots$
As written, this is an analytic statement, so issues of convergence must be considered.  (For instance, if $x$ is a real number, we need $|x| < 1$.)  But if it happens that some power of $x$ is equal zero, then so are all higher powers and the series is not infinite after all...With only a little work, one can make purely algebraic sense out of this.
