0

The exact statement I'm trying to prove is:

A square matrix $A$ is called \emph{nilpotent} if $A^k = 0$ for some $k > 0$. Prove that if $A$ is nilpotent, then $I + A$ is invertible.

I know how to prove this by brute force -- supplying an inverse and performing the multiplication on the left and right -- but I'm trying to understand an alternative proof by contradiction. My biggest concern is that I left off a case. (I'm uncomfortable, generally, working with a quantity "$k-1$" because I have to worry about the quantity either being negative or zero, which can cause issues.)

Here is my proof attempt.

Let $A$ be a nilpotent, $n \times n$ matrix in $\mathbb{F}$ of degree $k$. For the sake of contradiction, suppose that $I + A$ is not invertible. Hence, $\text{ker}($I + A$) $ is non-trivial by the rank-nullity theorem, hence there exists $v \in \mathbb{F}^n - \{0\}$ such that $(I+A)v = 0$. Then $v + Av = 0$, hence $Av = -v$. Applying $A^{k-1}$ yields $$ 0 = A^k v = - A^{k-1} v. $$ As $Av = -v$, one has $A^{k-1} v = (-1)^{k-1} v$ by induction. Indeed, if $k = 1$, the statement reduces to $Iv = v = (-1)^0 v$, which is true. If $A^{m-1} v = (-1)^{m-1} v$ for some $m \geq 1$, one has $$ A^m v = A(A^{m-1} v) = A((-1)^{m-1} v) = (-1)^{m-1} Av = (-1)^{m-1} ((-1)v) = (-1)^m v, $$ which closes the induction. Hence, $0 = (-1)^k v$. As $v \neq 0$ and $(-1)^k \neq 0$, this is a contradiction.

As I said, my biggest issue is the step where I apply $A^{k-1}$ to both sides of the equation. I believe my inductive proof works and the inductive step didn't require that $m \geq 2$ (in which case I'd need to prove a second base case), but I'm not fully confident in that. If $k = 1$ at any point, $A$ is just the zero matrix. I believe that is "allowed" but a rather trivial case because then $I + A = I$, which is certainly invertible. But if $k = 1$, $A^{k-1} = I$, which I can certainly apply rather seamlessly to both sides, so it doesn't seem to me this step requires special consideration. Am I wrong?

Mathematical Endeavors
  • 721
  • 4
    If the $k-1$ is bothering you, what's stopping you from applying $A^k$ to both sides instead? – Brian Moehring Feb 28 '23 at 01:17
  • Does this answer your question? Prove that $A+I$ is invertible if $A$ is nilpotent – Math1000 Feb 28 '23 at 15:33
  • That doesn't answer my question because I'm interested in this specific approach, as I mentioned. I know how to prove this fact by "brute force" by supplying an inverse, but I'm interested in a step of this specific proof. – Mathematical Endeavors Feb 28 '23 at 18:48
  • 1
    @MathematicalEndeavors This very quickly becomes a matter of subjective opinion, but I imagine you'll lose a lot of people if you continue to call the direct proof "brute force". We usually save that moniker for proofs that either require long, extensive case-work ("whack-a-mole") or lots of [ugly] calculations ("grunt work"). There is an academic value in searching for alternative proofs, but we can do so without effectively making value judgements. ---- Anyway, if you can answer the question I asked in my first comment; it's likely to help anyone who actually wants to answer. – Brian Moehring Mar 01 '23 at 22:54
  • I appreciate this comment. Thank you. You're also right: there's nothing preventing me from applying $A^k$ instead, and I probably should have done so. It does make me uneasy reusing $k$ in the context of an inductive proof where I've already fixed $k$, so perhaps I should change up the notation all together. – Mathematical Endeavors Mar 02 '23 at 04:31

1 Answers1

1

$A$ is nilpotent. So there exists a non-negative integer $k$ such that $A^k=0$. Now let $$B=I−A+A^2+\cdots+(−1)^{k−1}A^{k−1}$$ Clearly $$(I+A)B=B+AB=B+A(I−A+A^2+\dotsb)=I−(−1)^kA^k=I$$ Also $B(I+A)=I$. So $B$ is the inverse of $I+A$. Hence $I+A$ is invertible.

Sebastiano
  • 7,649