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!

For what it's worth I just want to mention that what is happening here is actually an instance of a more general result about rings. If $R$ is a ring then an element $a \in R$ is said to be nilpotent if there is $n \in \mathbb{N}$ such that $a^n = 0$.
In your question, the condition on your matrix that $A^2 = 0$ just means that it is nilpotent.
Now if $R$ is a ring with unity, then if $a \in R$ is nilpotent it can be proved that $1 - a$ is an invertible element (or unit) in the ring $R$, meaning that there is a $b \in R$ such that $(1 - a)b = b(1 - a) = 1$. From this you can then just change $a$ with $-a$ to also see that $1 + a$ is invertible.
I'm pretty sure that this is what Arturo had in mind by adding that exercise for bonus points for you, so I will not give the argument here. You can find it in this planetmath entry if you want to look at it. But I would suggest to you to first try it for yourself, for matrices at least.
