Mixing
Finding a Solution to Inequalities
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
asked Nov 21 '18 at 3:02
chandu1729
2,371821
add a comment |
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
asked Nov 21 '18 at 3:02
chandu1729
2,371821
add a comment |
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
asked Nov 21 '18 at 3:02
chandu1729
2,371821
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
asked Nov 21 '18 at 3:02
chandu1729
2,371821
asked Nov 21 '18 at 3:02
chandu1729
2,371821
asked Nov 21 '18 at 3:02
chandu1729
2,371821
asked Nov 21 '18 at 3:02
chandu1729
2,371821
asked Nov 21 '18 at 3:02
chandu1729
2,371821
2,371821
add a comment |
add a comment |
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.
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
add a comment |
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.
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
add a comment |
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.
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
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.
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
answered Nov 26 '18 at 22:10
Alex Ravsky
39.2k32180
39.2k32180
add a comment |
add a comment |
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