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For every large enough $m\in\Bbb N$ are there $c=\alpha m$ (for some fixed $\alpha>0$) square matrices $A_1,\dots,A_c$ that commute with each other with nonzero product ($\forall i,j\in\{1,\dots,t\}$ distinct, $A_iA_j=A_jA_i\neq 0$ holds) and satisfy $\forall i\in\{1,\dots,c\}$ $A_i^2=0$? What is smallest size of matrices (can they be $\beta m\times\beta m$ for some fixed $\beta>0$)?

I am ok with coefficients in any field with characteristic 0 or $O(2^{m^\gamma})$ for a fixed $\gamma>0$.

Similar problem: https://math.stackexchange.com/questions/1710735/commutating-nilpotent-operators.

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    $\begingroup$ I can certainly build $m$ matrices of size $\binom m2 + \binom m1+\binom m0=\binom{m+1}2$: Let the set $S$ consist of all subsets of $\{1,\ldots,m\}$ of size at most 2. Consider a vector space with one basis element for each element of $S$, where the basis element corresponding to $B$ is labelled $e_B$. Now define $A_ie_B$ to be 0 if $i\not\in B$ and $e_{B\setminus\{i\}}$ if $i\in B$. These matrices have the desired property. $\endgroup$ – Anthony Quas Mar 23 '16 at 23:31
  • $\begingroup$ @AnthonyQuas could you write down what the matrices look like for say $m=4$? $\endgroup$ – user76479 Mar 23 '16 at 23:50
  • $\begingroup$ There are 6+4+1 elements of $S$: Label them as $1\to\{1,2\}$, $2\to \{1,3\}$, $3\to \{1,4\}$, $4\to \{2,3\}$, $5\to\{2,4\}$, $6\to \{3,4\}$, $7\to\{1\}$, $8\to \{2\}$, $9\to\{3\}$, $10\to \{4\}$ and $11\to\emptyset$. Now $A_1$ maps $e_1$ to $e_8$ (since $\{1,2\}\setminus\{1\}=\{2\})$, $e_2$ to $e_9$, $e_3$ to $e_{10}$ and $e_7$ to $e_{11}$ (since $\{1\}\setminus\{1\}=\emptyset$). $A_1$ sends all other basis vectors to 0. Similarly $A_2$ maps $e_1$ to $e_7$, $e_4$ to $e_9$, $e_5$ to $e_{10}$ and $e_8$ to $e_{11}$ and all other basis vectors to 0. You can write down $A_3$ and $A_4$ similarly. $\endgroup$ – Anthony Quas Mar 24 '16 at 0:03
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It seems that here is a example of $k$ such transforms of the $(k+2)$-dimensional space. Let $(e_0,\dots,e_{m+1})$ be some fixed basis of the space.

The $i$th transform $\phi_i$ acts as follows: $\phi_i(e_0)=\phi_i(e_i)=0$, $\phi_i(e_j)=e_0$ for $1\leq j\leq m$, $j\neq i$, and $\phi_i(e_{m+1})=e_i$. Thus $\mathop{\rm Im} \phi_i=\langle e_0,e_i\rangle$, and thus $\phi_i\circ\phi_i=0$.

On the other hand, for every $i\neq j$, the transform $\phi_i\circ\phi_j$ maps $e_{m+1}$ to $e_0$, and all other vectors of the basis to $0$. THus all the remaining conditions are also satisfied.

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