We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.












2















We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.




Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.



There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?



Note: I already asked initaly question and did get an answer, so please answer just this.






linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics







share|cite|improve this question
























  • Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
    – Vincenzo
    2 days ago










  • @Vincenzo second one.
    – greedoid
    2 days ago










  • I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
    – Math1000
    2 days ago










  • Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
    – greedoid
    2 days ago






  • 1




    If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
    – Mike Earnest
    21 hours ago
















2















We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.




Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.



There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?



Note: I already asked initaly question and did get an answer, so please answer just this.






linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics







share|cite|improve this question
























  • Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
    – Vincenzo
    2 days ago










  • @Vincenzo second one.
    – greedoid
    2 days ago










  • I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
    – Math1000
    2 days ago










  • Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
    – greedoid
    2 days ago






  • 1




    If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
    – Mike Earnest
    21 hours ago














2












2








2








We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.




Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.



There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?



Note: I already asked initaly question and did get an answer, so please answer just this.






linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics







share|cite|improve this question
















We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.




Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.



There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?



Note: I already asked initaly question and did get an answer, so please answer just this.






linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics




linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago









greedoid

37.7k114794




37.7k114794












  • Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
    – Vincenzo
    2 days ago










  • @Vincenzo second one.
    – greedoid
    2 days ago










  • I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
    – Math1000
    2 days ago










  • Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
    – greedoid
    2 days ago






  • 1




    If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
    – Mike Earnest
    21 hours ago


















  • Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
    – Vincenzo
    2 days ago










  • @Vincenzo second one.
    – greedoid
    2 days ago










  • I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
    – Math1000
    2 days ago










  • Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
    – greedoid
    2 days ago






  • 1




    If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
    – Mike Earnest
    21 hours ago
















Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago




Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago












@Vincenzo second one.
– greedoid
2 days ago




@Vincenzo second one.
– greedoid
2 days ago












I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago




I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago












Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago




Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago




1




1




If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago




If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago










1 Answer
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oldest

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There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.






share|cite|improve this answer























  • Anyway, do you have a perhaps an idea how to atack this algebraicly
    – greedoid
    2 days ago










  • You think this is hopless try with linear algebra?
    – greedoid
    yesterday











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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

votes









1














There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.






share|cite|improve this answer























  • Anyway, do you have a perhaps an idea how to atack this algebraicly
    – greedoid
    2 days ago










  • You think this is hopless try with linear algebra?
    – greedoid
    yesterday
















1














There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.






share|cite|improve this answer























  • Anyway, do you have a perhaps an idea how to atack this algebraicly
    – greedoid
    2 days ago










  • You think this is hopless try with linear algebra?
    – greedoid
    yesterday














1












1








1






There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.






share|cite|improve this answer














There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago









greedoid

37.7k114794




37.7k114794










answered 2 days ago









Alex Ravsky

38.8k32079




38.8k32079












  • Anyway, do you have a perhaps an idea how to atack this algebraicly
    – greedoid
    2 days ago










  • You think this is hopless try with linear algebra?
    – greedoid
    yesterday


















  • Anyway, do you have a perhaps an idea how to atack this algebraicly
    – greedoid
    2 days ago










  • You think this is hopless try with linear algebra?
    – greedoid
    yesterday
















Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago




Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago












You think this is hopless try with linear algebra?
– greedoid
yesterday




You think this is hopless try with linear algebra?
– greedoid
yesterday


















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