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

Question

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?

Answer 1

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 $k<n$, 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 $AB\neq BA$.

Answer 2

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 ?

Answer 3

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!