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LSpice
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Does there always exist such a matrix satisfying certain tracial conditions

Given odd integers $0<a<b$, I want to know if there exists an $n$ by $n$ real valued square matrix $M$ such that $$ M_{ij} = M_{ji} \quad \forall i,j \in \{1,2\dots n\}$$ $$ \sum_{i=1}^n M_{ij} = 0\quad \forall j \in \{1,2\dots n\} $$ $$\operatorname{Tr}(M^a)\operatorname{Tr}(M^b) = \sum_{i=1}^n \lambda_i^a \sum_{i=1}^n\lambda_i^b < 0 $$ for some finite $n$.

If there always exists such an $M$, this would resolve a case which may lead to improving a result of    this paperLocally common graphs by Csóka, Hubai, and Lovász. In the paper's language, I am trying to prove that there always exists a balanced graphon $U$ such that $t(C_a,U)t(C_b,U)<0$. Proposition 4.3 proves a similar special case, where the two cycles are connected. Since they ignored the disconnected case, I suspect my problem maybemay be trivial,. I am not too familiar with manipulating the trace function, so I'm not sure.

Does there always exist such a matrix

Given odd integers $0<a<b$, I want to know if there exists an $n$ by $n$ real valued square matrix $M$ such that $$ M_{ij} = M_{ji} \quad \forall i,j \in \{1,2\dots n\}$$ $$ \sum_{i=1}^n M_{ij} = 0\quad \forall j \in \{1,2\dots n\} $$ $$\operatorname{Tr}(M^a)\operatorname{Tr}(M^b) = \sum_{i=1}^n \lambda_i^a \sum_{i=1}^n\lambda_i^b < 0 $$ for some finite $n$.

If there always exists such an $M$, this would resolve a case which may lead to improving a result of  this paper by Lovász. In the paper's language, I am trying to prove that there always exists a balanced graphon $U$ such that $t(C_a,U)t(C_b,U)<0$. Proposition 4.3 proves a similar special case, where the two cycles are connected. Since they ignored the disconnected case, I suspect my problem maybe trivial, I am not too familiar with manipulating the trace function, so I'm not sure.

Does there always exist a matrix satisfying certain tracial conditions

Given odd integers $0<a<b$, I want to know if there exists an $n$ by $n$ real valued square matrix $M$ such that $$ M_{ij} = M_{ji} \quad \forall i,j \in \{1,2\dots n\}$$ $$ \sum_{i=1}^n M_{ij} = 0\quad \forall j \in \{1,2\dots n\} $$ $$\operatorname{Tr}(M^a)\operatorname{Tr}(M^b) = \sum_{i=1}^n \lambda_i^a \sum_{i=1}^n\lambda_i^b < 0 $$ for some finite $n$.

If there always exists such an $M$, this would resolve a case which may lead to improving a result of  Locally common graphs by Csóka, Hubai, and Lovász. In the paper's language, I am trying to prove that there always exists a balanced graphon $U$ such that $t(C_a,U)t(C_b,U)<0$. Proposition 4.3 proves a similar special case, where the two cycles are connected. Since they ignored the disconnected case, I suspect my problem may be trivial. I am not too familiar with manipulating the trace function, so I'm not sure.

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Zach Hunter
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Does there always exist such a matrix

Given odd integers $0<a<b$, I want to know if there exists an $n$ by $n$ real valued square matrix $M$ such that $$ M_{ij} = M_{ji} \quad \forall i,j \in \{1,2\dots n\}$$ $$ \sum_{i=1}^n M_{ij} = 0\quad \forall j \in \{1,2\dots n\} $$ $$\operatorname{Tr}(M^a)\operatorname{Tr}(M^b) = \sum_{i=1}^n \lambda_i^a \sum_{i=1}^n\lambda_i^b < 0 $$ for some finite $n$.

If there always exists such an $M$, this would resolve a case which may lead to improving a result of this paper by Lovász. In the paper's language, I am trying to prove that there always exists a balanced graphon $U$ such that $t(C_a,U)t(C_b,U)<0$. Proposition 4.3 proves a similar special case, where the two cycles are connected. Since they ignored the disconnected case, I suspect my problem maybe trivial, I am not too familiar with manipulating the trace function, so I'm not sure.