# Nilpotent, upper triangular sum and product of matrices

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?

• It seems that previously I've made an error in the question to having written that index of nilpotency ( i.e. such minimal $k,m$ for $A+B$ and $AB$ that $(A+B)^k=0$ and $(AB)^m=0$ - I'm not sure whether always $m=k$ ) should be greater than $1$ - it's an obvious case , more proper assumption is that they should be greater than $2$ Commented Jul 12, 2023 at 10:18
• With your notation, not always $m=k$. Take $3\times 3$ matrices $A=(a_{i,j}),B=(b_{i,j})$ all zeroes except $a_{1,3}=b_{1,2} = b_{2,1} = 1$. Then $AB$ is the zero matrix, but $A+B$ has index 3. Commented Jul 12, 2023 at 12:12

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.

• And what about the index of nilpotency for $A+B$ and $AB$. Is it important or not? Commented Jul 12, 2023 at 10:03
• No, you see in my proof that there is no condition on the indices of nilpotence at all. Commented Jul 12, 2023 at 11:01
• Hmm, could you give me an example of $A+B$ and $AB$ with index nilpotency equal to $2$? ( simpler would be to show $A$ and $B$) because I have not found such ones. Exclude, of course, zero matrices. Commented Jul 12, 2023 at 11:11
• You want an example of two nilpotent matrices $A$ and $B$ such that $A+B$ and $AB$ are nilpotent of index $=2$ ? If so, this is a very different question from the original one. Commented Jul 12, 2023 at 11:22
• Take $$A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \quad \text{and} \quad B=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}$$ Then $A+B$ and $AB$ are both nilpotent of index $=2$. Commented Jul 12, 2023 at 11:53

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.

• And what about the index of nilpotency for $A+B$ and $AB$. Is it important or not? Commented Jul 12, 2023 at 10:03
• It does not seem to be relevant, but in the upper-diagonal case with diagonals zeroes, the nilpotency index of $AB$ is always a lower bound on the indexes of $A$ and $B$. If $A,B$ are of size $n$, upper diagonal and with zeroes on the diagonal, then $n$ is an upper bound. Commented Jul 12, 2023 at 12:09
• Thank you for the answer, I've upvoted it (+1). It is similar to that of TheSilverDoe so I could only mark one as accepted, but your answer is also very good. Commented Jul 12, 2023 at 12:31