Nilpotent, upper triangular sum and product of matrices

Question

In the MSE question it was claimed that if matrices $AB$ and $A+B$ are nilpotent then $A,B$ are nilpotent. However generally the claim is false - user1551 has found quickly counterexample.

I wonder whether the claim would be true when we would add one additional condition for $AB$ and $A+B$, namely that they are upper triangular.

Possibly we should also declare that dimension of matrices must be greater than $2 \times 2$, although maybe it's not enough, stronger condition seems to be that index of nilpotency for $AB$ and $A+B$ is greater than $2$.

Could someone confirm or reject my assumptions?

Answer 1

Let's prove that the result is true is $A$ anb $B$ are upper triangular : for this, let $A$ and $B$ be two upper-triangular matrices such that $A+B$ and $AB$ are nilpotent.

Let $(a_i)_{1 \leq i \leq n}$ denote the diagonal elements of $A$, and $(b_i)_{1 \leq i \leq n}$ the diagonal elements of $B$.

• $A+B$ is upper triangular with diagonal elements $(a_i+b_i)_{1 \leq i \leq n}$ : since it is nilpotent, then $a_i + b_i = 0$ for every $i=1, ..., n$, i.e. $a_i=-b_i$.

• But $AB$ is also upper triangular, with diagonal elements $(a_ib_i)_{1 \leq i \leq n}$ : since it is nilpotent, then $a_i b_i = 0$ for every $i=1, ..., n$, i.e. (since $a_i=-b_i$), one has $-a_i^2=$, i.e. $a_i=b_i=0$.

So $A$ and $B$ are upper triangular with zero diagonal elements : so they are nilpotent.

Answer 2

The product and sum of upper-triangular matrices is upper-triangular. An upper-triangular matrix $A$ is nilpotent if and only if all of its diagonal entries are zero, since the diagonal entries of $A^n$ are the diagonal entries of $A$ to the power of $n$, and if all the diagonal entries of an upper-diagonal matrix are zero, then some power of $A$ will eventually be zero (see this math stack exchange post).

If $A,B$ are upper-diagonal matrices, then $AB$, $A+B$ are upper diagonal. If $a_i$ are the elements on the diagonal of $A$, $b_i$ of the diagonal of $B$, then for $AB$ to be nilpotent, $a_ib_i = 0$ for all $i$, while for $A+B$ to be nilpotent $a_i+b_i=0$ for each $i$. This can only happen if $a_i=b_i=0$ for each $i$, hence the diagonals of $A,B$ are zero and so by the previous discussion they are nilpotent.