# Commutating nilpotent operators

Is there good examples of collection of nilpotent operators that commute with themselves?

Is there a good reference for a collection commutative nilpotent operators that commute with themselves or commutative zero divisors?

Specifically for large enough $m\in\Bbb N$ I am looking for $t=\alpha m$ for some fixed $\alpha>0$ matrices $A_1,\dots,A_t$ that are commutative ( $\forall i,j\in\{1,\dots,t\}$ $A_iA_j=A_jA_i$ holds) and satisfy one of following two:

(1) For any $i\in\{1,\dots,t\}$ $A_i^2=0$. What is smallest size of matrices (can they be $\beta m\times\beta m$ for some fixed $\beta>0$)?

(2) $\forall i_1,i_2,i_3,\dots\in\{1,\dots,t\}$ we have $Tr(A_{i_1}A_{i_2}A_{i_3}\dots)\neq0\iff {i_1}\neq {i_2}\wedge {i_2}\neq {i_3}\wedge {i_3}\neq {i_1}\dots$ holds. Can matrices also be of size $\beta m\times\beta m$ for some fixed $\beta>0$?

• Commutative with what ? Among themselves or with other operators? – DonAntonio Mar 23 '16 at 20:26
• With other operators. – user257494 Mar 23 '16 at 20:28
• Look inside some commutative ring, for example $k[x]/(x^n)$, or $k[x,y] / (xy)$. Your question is very broad. Maybe you can ask for something more specific? (Put more conditions on your operators.) – Lorenzo Mar 23 '16 at 20:29
• The only operators that are commutative with all other operators are the scalar ones, and the only scalar nilpotent operators is the zero one. – DonAntonio Mar 23 '16 at 20:29
• I am looking for matrix examples. – user257494 Mar 23 '16 at 20:29

## 1 Answer

Here is a theorem that is relevant to your question: A Lie sub algebra of $End(V)$ (V finite dimensional) consisting of nilpotent operators is simultaneously strictly upper triangularizable. I mean that you can pick a basis so that they all are zero on and below the diagonal. (This is Engels theorem. A nice reference is Serre's Lie algebras and Lie groups.)

(The condition on commutativity means that these form a Lie sub algebra.)

I'm still a little confused about what specifically you are asking, but you can probably play around with this a bit to produce a result.

Concrete translation of a specific case:

Let $A_i$ be a collection of commuting, nilpotent operators on a finite dimensional vector space $V$. Then there is a basis $v_i$ for $V$ in which each $A_i$ is represented by a strictly upper triangular matrix.

• @DietrichBurde The lie algebra generated by them contains a diagonal matrix, which is not nilpotent. What I mean is that all of the operators in the Lie algebra are nilpotent. – Lorenzo Mar 23 '16 at 20:48
• No, he means Engel's theorem, and yes, this assumption is sufficient ; your counterexample is incorrect because the Lie subalgebra generated by nilpotent elements will not in general consist in nilpotent element (and he did specify that all the elements of the Lie subalgebra should be nilpotent, which always happens if it is generated by commuting nilpotent elements). – Captain Lama Mar 23 '16 at 20:48
• @AreaMan Could you be more explicit and matrix theoretic since I am looking for concrete examples? – user257494 Mar 23 '16 at 20:55
• when you say the Lie algebra generated by $(A_i)_{i \in \{1 \ldots t\}}$ where $A_i A_j = A_j A_i$, you mean matrices of the form $\sum_{i=1}^t c_i A_i$ and that's all ? en.wikipedia.org/wiki/Lie_algebra#Three_dimensions – reuns Mar 23 '16 at 20:57
• @AreaMan Yes, you are right. Serre has this result under the name "Engel". Usually Engel's theorem is that a Lie algebra is nilpotent if and only if all adjoint operators are nilpotent. – Dietrich Burde Mar 23 '16 at 21:12