# Let $A$ and $B$ be two nilpotent matrices. Prove that $A+B$ is nilpotent

Let $$A$$ and $$B$$ be two nilpotent matrices. Prove that $$A+B$$ is nilpotent.

I'm new to the study of matrices, I already proved that $$AB$$ is nilpotent too (that was the previous question) but I have no clue about solving this one. Plus, I was wondering if I have to prove that it exists an $$n$$ such that $$(A+B)^n=0$$ or such that $$A^n+B^n=0$$.

Any help?

EDIT:

I just know that: $$(A+B)^k=(A+B)(A+B)...(A+B)$$ $$k$$ times, so I need to prove that $$(A+B)=0$$

Is this enough to prove what I need, right?

• $A=\pmatrix{0&1\\0&0}$ and $B=\pmatrix{0&0\\1&0}$ are nilpotent. Commented Nov 1, 2019 at 14:27
• $(A+B)^n \neq A^n + B^n$. If you are wanting to show that $C$ is nilpotent that means you want to show that there is some $n$ such that $C^n$ is nilpotent. Replacing $C$ by $(A+B)$ that says that if you want to show that $(A+B)$ is nilpotent that you want to find some $n$ such that $(A+B)^n$ is nilpotent. Commented Nov 1, 2019 at 14:28
• If $A$ and $B$ commute, then this is true by the binomial theorem. Otherwise, there are counterexamples.
– lhf
Commented Nov 1, 2019 at 14:31

Take for example the nilpotent matrices

$$A=\begin{pmatrix}2&-1\\4&-2\end{pmatrix}\;,\;\;\;B=\begin{pmatrix}0&1\\0&0\end{pmatrix}\implies\text{ what is}\;\;A+B \;?$$

Why that sum can't be nilpotent ?

• Sorry, I don't get it. I mean, I know what matrix $A+B$ is, but how is this related to the fact of being nilpotent or not? Commented Nov 1, 2019 at 14:53
• @Schiele $A+B$ is invertible and an invertible matrix cannot be nilpotent. Commented Nov 1, 2019 at 15:15

Assume that there are $$m,n\in\mathbb{N}$$ such that $$A^m=0,B^n=0.$$

Then $$(A+B)^{m+n}=\sum_{k=0}^{m+n} C_{m+n}^k A^{m+n-k}B^{k}$$ provided AB=BA.

When $$k, we know $$m+n-k>m$$, which means $$A^{m+n-k}=0$$. And when $$k\geq n$$, we know $$B^k=0.$$ That is $$(A+B)^{m+n}=0,$$ which means A+B is nilpotent.

But I have no idea how to answer it if $$AB\neq BA$$.

• Thank you very much! But I didn't understand why when $k>n$ we know that $B^k=0$? Commented Nov 1, 2019 at 15:16
• @Schiele Ling says in the first line that B^n=0. Commented Nov 1, 2019 at 16:00

I think you're missing a hypothesis: the group of the matrices has to be commutative! Otherwise the thesis doesn't hold:

let $$A= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ and $$B= \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$

We have that $$A^2=0$$ and $$B^2=0$$, but $$(A+B)^2= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ which is definitely not the zero matrix!