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!

The minus sign is not an obstacle: If $AB = -I$, then $A(-B) = -(AB) = -(-I) = I$. So in fact, if $A^2 = 0$, then $(A+I)(I-A) = A - A^2 + I - A = I$, so $A+I$ is invertible, as your first professor noted.
The error in the second argument is the following: It is true that if $B\mathbf{x}=\mathbf{0}$ has a nontrivial solution, then $CB\mathbf{x}=\mathbf{0}$ has a nontrivial solution. Thus, if $B$ is not invertible, then $CB$ is not invertible. But that is not what was argued. What was argued instead was that since $CB\mathbf{x}=\mathbf{0}$ has a nontrivial solution, then it follows that $B\mathbf{x}=\mathbf{0}$ has a nontrivial solution (with $B=A+I$ and $C=A$). This argument is incorrect: you can always take $C=0$, and that would mean that no matrix is invertible. 
It is certainly true that if $A$ is not invertible, then no multiple of $A$ is invertible (so for every $C$, neither $CA$ nor $AC$ are invertible); so you can deduce that $A(A+I)$ is not invertible. This does not prove that $A+I$ is not invertible, however, which is what you wanted to show.
Now, for bonus points, show that if $A$ is an $n\times n$ matrix and $A^k=0$ for some positive integer $k$, then $A+\lambda I_n$ is invertible for any nonzero $\lambda$. 
Added: For bonus bonus points, explain why the argument would break down if we replace $\lambda I_n$ with an arbitrary invertible matrix $B$. 
