Finding a Solution to Inequalities












1














I am trying to tackle the following problem:



Given matrix $alpha=(alpha_{ij})$ such that all the entries are non-negative and $alpha_{ij}alpha_{ji}leq 1$ for all $i,j$ and $alpha_{ii}=1$ for all $i$. Find a vector $c=(c_1,c_2, cdots , c_n)geq 0$ such that $alpha_{ji}leq frac{c_i}{c_j} leq frac{1}{alpha_{ij}}$. (Clearly, this can have many solutions and we are just interested in one possible solution.)



I tried by assuming $c_1=1$ and use terms like $max_{i}alpha_{ij}$ but none of the attempts are successful. I would be thankful for any help!






combinatorics matrices inequality reference-request







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    1














    I am trying to tackle the following problem:



    Given matrix $alpha=(alpha_{ij})$ such that all the entries are non-negative and $alpha_{ij}alpha_{ji}leq 1$ for all $i,j$ and $alpha_{ii}=1$ for all $i$. Find a vector $c=(c_1,c_2, cdots , c_n)geq 0$ such that $alpha_{ji}leq frac{c_i}{c_j} leq frac{1}{alpha_{ij}}$. (Clearly, this can have many solutions and we are just interested in one possible solution.)



    I tried by assuming $c_1=1$ and use terms like $max_{i}alpha_{ij}$ but none of the attempts are successful. I would be thankful for any help!






    combinatorics matrices inequality reference-request







    share|cite|improve this question

























      1












      1








      1







      I am trying to tackle the following problem:



      Given matrix $alpha=(alpha_{ij})$ such that all the entries are non-negative and $alpha_{ij}alpha_{ji}leq 1$ for all $i,j$ and $alpha_{ii}=1$ for all $i$. Find a vector $c=(c_1,c_2, cdots , c_n)geq 0$ such that $alpha_{ji}leq frac{c_i}{c_j} leq frac{1}{alpha_{ij}}$. (Clearly, this can have many solutions and we are just interested in one possible solution.)



      I tried by assuming $c_1=1$ and use terms like $max_{i}alpha_{ij}$ but none of the attempts are successful. I would be thankful for any help!






      combinatorics matrices inequality reference-request







      share|cite|improve this question













      I am trying to tackle the following problem:



      Given matrix $alpha=(alpha_{ij})$ such that all the entries are non-negative and $alpha_{ij}alpha_{ji}leq 1$ for all $i,j$ and $alpha_{ii}=1$ for all $i$. Find a vector $c=(c_1,c_2, cdots , c_n)geq 0$ such that $alpha_{ji}leq frac{c_i}{c_j} leq frac{1}{alpha_{ij}}$. (Clearly, this can have many solutions and we are just interested in one possible solution.)



      I tried by assuming $c_1=1$ and use terms like $max_{i}alpha_{ij}$ but none of the attempts are successful. I would be thankful for any help!






      combinatorics matrices inequality reference-request




      combinatorics matrices inequality reference-request






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      asked Nov 21 '18 at 3:02









      chandu1729

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          Unfortunately, there is no such vector $c$ in general. A counterexample is the following. Let $alpha$ be $3times 3$ matrix such that $alpha_{13}=alpha_{32}=2$, $alpha_{23}=alpha_{31}=frac 12$, and all remaining $alpha_{ij}$ equal to $1$. Assume that $c$ is the required vector. Remark that since each $c_i$ can be placed in a denominator, all $c_i$ are non-negative. We have $1=alpha_{21}lefrac{c_1}{c_2}lefrac {1}{alpha_{12}}=1$, so $c_1=c_2=c’$. Also we have $alpha_{32}c_3le c’$ and $alpha_{13}c’le c_3$. That is $2c_3le c’$ and $2c’le c_3$. Hence $2c_3le c’le frac 12 c_3$, a contradiction.






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            1 Answer
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            Unfortunately, there is no such vector $c$ in general. A counterexample is the following. Let $alpha$ be $3times 3$ matrix such that $alpha_{13}=alpha_{32}=2$, $alpha_{23}=alpha_{31}=frac 12$, and all remaining $alpha_{ij}$ equal to $1$. Assume that $c$ is the required vector. Remark that since each $c_i$ can be placed in a denominator, all $c_i$ are non-negative. We have $1=alpha_{21}lefrac{c_1}{c_2}lefrac {1}{alpha_{12}}=1$, so $c_1=c_2=c’$. Also we have $alpha_{32}c_3le c’$ and $alpha_{13}c’le c_3$. That is $2c_3le c’$ and $2c’le c_3$. Hence $2c_3le c’le frac 12 c_3$, a contradiction.






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              Unfortunately, there is no such vector $c$ in general. A counterexample is the following. Let $alpha$ be $3times 3$ matrix such that $alpha_{13}=alpha_{32}=2$, $alpha_{23}=alpha_{31}=frac 12$, and all remaining $alpha_{ij}$ equal to $1$. Assume that $c$ is the required vector. Remark that since each $c_i$ can be placed in a denominator, all $c_i$ are non-negative. We have $1=alpha_{21}lefrac{c_1}{c_2}lefrac {1}{alpha_{12}}=1$, so $c_1=c_2=c’$. Also we have $alpha_{32}c_3le c’$ and $alpha_{13}c’le c_3$. That is $2c_3le c’$ and $2c’le c_3$. Hence $2c_3le c’le frac 12 c_3$, a contradiction.






              share|cite|improve this answer
























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                Unfortunately, there is no such vector $c$ in general. A counterexample is the following. Let $alpha$ be $3times 3$ matrix such that $alpha_{13}=alpha_{32}=2$, $alpha_{23}=alpha_{31}=frac 12$, and all remaining $alpha_{ij}$ equal to $1$. Assume that $c$ is the required vector. Remark that since each $c_i$ can be placed in a denominator, all $c_i$ are non-negative. We have $1=alpha_{21}lefrac{c_1}{c_2}lefrac {1}{alpha_{12}}=1$, so $c_1=c_2=c’$. Also we have $alpha_{32}c_3le c’$ and $alpha_{13}c’le c_3$. That is $2c_3le c’$ and $2c’le c_3$. Hence $2c_3le c’le frac 12 c_3$, a contradiction.






                share|cite|improve this answer












                Unfortunately, there is no such vector $c$ in general. A counterexample is the following. Let $alpha$ be $3times 3$ matrix such that $alpha_{13}=alpha_{32}=2$, $alpha_{23}=alpha_{31}=frac 12$, and all remaining $alpha_{ij}$ equal to $1$. Assume that $c$ is the required vector. Remark that since each $c_i$ can be placed in a denominator, all $c_i$ are non-negative. We have $1=alpha_{21}lefrac{c_1}{c_2}lefrac {1}{alpha_{12}}=1$, so $c_1=c_2=c’$. Also we have $alpha_{32}c_3le c’$ and $alpha_{13}c’le c_3$. That is $2c_3le c’$ and $2c’le c_3$. Hence $2c_3le c’le frac 12 c_3$, a contradiction.







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                answered Nov 26 '18 at 22:10









                Alex Ravsky

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