Let $A$ and $B$ be nilpotent $n \times n$ matrices with $AB = BA$. Show that $A + B$ is also nilpotent.

Question

Because $A$ and $B$ are nilpotent we know that $A^n$ and $B^n$ are $0$.

I guess we can rewrite $(A+B)^n$ as $\sum_{k=0}^n {n \choose k}A^kB^{n-k}$ and then use $AB = BA$ to rewrite this in a way that shows $(A+B)^n$ is equal to $0$.

It's already clear that the first and last term are $0$ so we can write $(A+B)^n = \sum_{k=1}^{n-1} {n \choose k} A^kB^{n-k}$ but I have no clue what to do with the other terms.

Also, your expression $(A + B)^n = \sum_{k = 0}^n \binom n k A^k B^{n - k}$ already used that $A B = B A$. — LSpice, May 13 '23 at 13:57

score 2 · Answer 1 · answered May 13 '23 at 13:56

2

Since $A$ and $B$ commute we can write $$(A+B)^{2n-1}=\sum_{k=0}^{2n-1} {2n-1 \choose k}A^kB^{2n-1-k}.$$ Either $k$ or $2n-1-k$ is at least $n$ and so each term of the summation is the zero matrix.

answered May 13 '23 at 13:56

user1172706

1,405

Let $A$ and $B$ be nilpotent $n \times n$ matrices with $AB = BA$. Show that $A + B$ is also nilpotent.

1 Answers1