Mixing
A notion of adjacency-matrix symmetry?
$begingroup$
I have a set of adjacency matrices that have a certain property, and I am trying to figure out what features of the adjacency matrix deliver that property and whether this property has a name.
Here is the property: Let $A$ be an adjacency matrix for an undirected graph. Such a matrix has Property $X$ if all the diagonal elements of $A$ are the same (which of course they are trivially), all the diagonal elements of $A^2$ are the same, all the diagonal elements of $A^3$ are the same, and so on for all powers of $A$.
Another way of describing this property is that for each node $i$, there are the same number of walks of length $k$ from node $i$ to itself for all $k$.
Thanks!
linear-algebra matrices graph-theory
asked Jan 18 at 23:10
MikeMike
368110
$endgroup$
add a comment |
$begingroup$
I have a set of adjacency matrices that have a certain property, and I am trying to figure out what features of the adjacency matrix deliver that property and whether this property has a name.
Here is the property: Let $A$ be an adjacency matrix for an undirected graph. Such a matrix has Property $X$ if all the diagonal elements of $A$ are the same (which of course they are trivially), all the diagonal elements of $A^2$ are the same, all the diagonal elements of $A^3$ are the same, and so on for all powers of $A$.
Another way of describing this property is that for each node $i$, there are the same number of walks of length $k$ from node $i$ to itself for all $k$.
Thanks!
linear-algebra matrices graph-theory
asked Jan 18 at 23:10
MikeMike
368110
$endgroup$
$begingroup$
"... for all $i$'', not "for all $k$", I suppose?
$endgroup$
– W-t-P
Jan 19 at 7:47
add a comment |
$begingroup$
I have a set of adjacency matrices that have a certain property, and I am trying to figure out what features of the adjacency matrix deliver that property and whether this property has a name.
Here is the property: Let $A$ be an adjacency matrix for an undirected graph. Such a matrix has Property $X$ if all the diagonal elements of $A$ are the same (which of course they are trivially), all the diagonal elements of $A^2$ are the same, all the diagonal elements of $A^3$ are the same, and so on for all powers of $A$.
Another way of describing this property is that for each node $i$, there are the same number of walks of length $k$ from node $i$ to itself for all $k$.
Thanks!
linear-algebra matrices graph-theory
asked Jan 18 at 23:10
MikeMike
368110
$endgroup$
I have a set of adjacency matrices that have a certain property, and I am trying to figure out what features of the adjacency matrix deliver that property and whether this property has a name.
Here is the property: Let $A$ be an adjacency matrix for an undirected graph. Such a matrix has Property $X$ if all the diagonal elements of $A$ are the same (which of course they are trivially), all the diagonal elements of $A^2$ are the same, all the diagonal elements of $A^3$ are the same, and so on for all powers of $A$.
Another way of describing this property is that for each node $i$, there are the same number of walks of length $k$ from node $i$ to itself for all $k$.
Thanks!
linear-algebra matrices graph-theory
linear-algebra matrices graph-theory
asked Jan 18 at 23:10
MikeMike
368110
asked Jan 18 at 23:10
MikeMike
368110
edited Jan 18 at 23:56
Mike
asked Jan 18 at 23:10
MikeMike
368110
asked Jan 18 at 23:10
MikeMike
368110
asked Jan 18 at 23:10
MikeMike
368110
368110
$begingroup$
"... for all $i$'', not "for all $k$", I suppose?
$endgroup$
– W-t-P
Jan 19 at 7:47
add a comment |
$begingroup$
"... for all $i$'', not "for all $k$", I suppose?
$endgroup$
– W-t-P
Jan 19 at 7:47
$begingroup$
"... for all $i$'', not "for all $k$", I suppose?
$endgroup$
– W-t-P
Jan 19 at 7:47
$begingroup$
"... for all $i$'', not "for all $k$", I suppose?
$endgroup$
– W-t-P
Jan 19 at 7:47
add a comment |
1 Answer
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The usual name for this property is walk regular. and google will quickly lead to a lot of information. Clearly any vertex-transitive graph is walk regular, more generally any graph that is a union of classes from an association scheme. Connected regular graphs with exactly four eigenvalues are another class. I do not know of a characterization of walk-regular graphs that is much better than the definition.
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The usual name for this property is walk regular. and google will quickly lead to a lot of information. Clearly any vertex-transitive graph is walk regular, more generally any graph that is a union of classes from an association scheme. Connected regular graphs with exactly four eigenvalues are another class. I do not know of a characterization of walk-regular graphs that is much better than the definition.
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
$endgroup$
add a comment |
$begingroup$
The usual name for this property is walk regular. and google will quickly lead to a lot of information. Clearly any vertex-transitive graph is walk regular, more generally any graph that is a union of classes from an association scheme. Connected regular graphs with exactly four eigenvalues are another class. I do not know of a characterization of walk-regular graphs that is much better than the definition.
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
$endgroup$
add a comment |
$begingroup$
The usual name for this property is walk regular. and google will quickly lead to a lot of information. Clearly any vertex-transitive graph is walk regular, more generally any graph that is a union of classes from an association scheme. Connected regular graphs with exactly four eigenvalues are another class. I do not know of a characterization of walk-regular graphs that is much better than the definition.
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
$endgroup$
The usual name for this property is walk regular. and google will quickly lead to a lot of information. Clearly any vertex-transitive graph is walk regular, more generally any graph that is a union of classes from an association scheme. Connected regular graphs with exactly four eigenvalues are another class. I do not know of a characterization of walk-regular graphs that is much better than the definition.
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
answered Jan 19 at 16:30
Chris GodsilChris Godsil
11.7k21635
11.7k21635
add a comment |
add a comment |
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$begingroup$
"... for all $i$'', not "for all $k$", I suppose?
$endgroup$
– W-t-P
Jan 19 at 7:47