If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible?

Solution 1:

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 $B$.

Solution 2:

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.