From Serge Lang's linear algebra textbook:
A square matrix $A$ is said to be nilpotent if $A^r=O$ for some integer $r \geq 1$. Let $A$, $B$ be nilpotent matrices, of the same size, and assume $AB$=$BA$. Show that $AB$ and $A+B$ are nilpotent.
By the way, $O$ denotes a null vector.
Simple explanation would be that, if matrix is symmetric and nilpotent, multiplication by itself will always lead to zero vector.
From what I know, matrices are closed under addition and multiplication. Therefore shouldn't linear combination of nilpotent matrix form another nilpotent matrix?
Furthermore, according to Wolfram Alpha:
Since $n×n$ matrices form an Abelian group under addition, n×n matrices form a ring.
However, matrix multiplication is not, in general, commutative (although it is commutative if $A$ and $B$ are diagonal and of the same dimension).
According to my interpretation of the text above, matrix addition is commutative (since commutativity is fundamental rule of abelian groups).
But what's important, is that if $AB=BA$, then the matrices must be diagonal. Let matrix $A$ be diagonal matrix with diagonal elements $a_{ij}$, then $A^k$ where $k \geq 1$ will give $a_{ij}^k$ (so that, all diagonal elements of matrix $A$ will be raised to the power $k$).
Now according to the paragraph above, I assume that $A$ and $B$ are $O$, since $n^{k}=0$ if $n=0$. But if $n \geq 1$ then $n^{k} \neq 0$ (if $k$ is positive integer, of course).
Finally, it gives that $AB=OO=O$ and $A+B=O+O=O$, so that they are nilpotent?
My "proof" might have fundamental mistakes, therefore what are better ways of proving this?
Thank you!
