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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?

Widawensen
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  • 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 $2$ – Widawensen Jul 12 '23 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. – nelynx Jul 12 '23 at 12:12

2 Answers2

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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.

nelynx
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  • And what about the index of nilpotency for $A+B$ and $AB$. Is it important or not? – Widawensen Jul 12 '23 at 10:03
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    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. – nelynx Jul 12 '23 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. – Widawensen Jul 12 '23 at 12:31
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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.

TheSilverDoe
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  • And what about the index of nilpotency for $A+B$ and $AB$. Is it important or not? – Widawensen Jul 12 '23 at 10:03
  • No, you see in my proof that there is no condition on the indices of nilpotence at all. – TheSilverDoe Jul 12 '23 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. – Widawensen Jul 12 '23 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. – TheSilverDoe Jul 12 '23 at 11:22
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    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 $=2$. – TheSilverDoe Jul 12 '23 at 11:53
  • I was originally asking about index of nilpotency. But your example shows that my intuitive restrictions for this index were unjustified. Thank you for the answer. – Widawensen Jul 12 '23 at 12:26