Eigenvalues of an adjacency matrix
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I am trying to prove or disprove the following:
Let $G$ be a non-bipartite AND connected $k$-regular graph where the size of the second-largest eigenvalue of the adjacency matrix $A_G$ of $G$ is $lambda$ (which as $G$ is connected is always strictly less than $k$). Then every eigenvalue of $A_G$, except for one, has absolute value no larger than $|lambda|$.
Of course, the shaded above would not be true if $G$ is allowed to be bipartite; if $G$ is $k$-regular bipartite the eigenvalues of $A_G$ are symmetric around 0, so $-k$ is an eigenvalue of $A_G$ for any $k$-regular bipartite $G$. I am wondering if there is a $k$-regular connected graph $G$ that is not bipartite that has smallest eigenvalue in the interval $(-k,-|lambda|]$, where $lambda$ is as above the 2nd-largest eigenvalue of $A_G$.
I have been looking around the internet for a solution either way, meanwhile my matrix analysis is rusty and I know we have some really smart people here. Anyone have any insight into this?
Thanks!
combinatorics graph-theory matrix-analysis
asked Nov 18 at 22:12
Mike
2,624211
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up vote
0
down vote
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I am trying to prove or disprove the following:
Let $G$ be a non-bipartite AND connected $k$-regular graph where the size of the second-largest eigenvalue of the adjacency matrix $A_G$ of $G$ is $lambda$ (which as $G$ is connected is always strictly less than $k$). Then every eigenvalue of $A_G$, except for one, has absolute value no larger than $|lambda|$.
Of course, the shaded above would not be true if $G$ is allowed to be bipartite; if $G$ is $k$-regular bipartite the eigenvalues of $A_G$ are symmetric around 0, so $-k$ is an eigenvalue of $A_G$ for any $k$-regular bipartite $G$. I am wondering if there is a $k$-regular connected graph $G$ that is not bipartite that has smallest eigenvalue in the interval $(-k,-|lambda|]$, where $lambda$ is as above the 2nd-largest eigenvalue of $A_G$.
I have been looking around the internet for a solution either way, meanwhile my matrix analysis is rusty and I know we have some really smart people here. Anyone have any insight into this?
Thanks!
combinatorics graph-theory matrix-analysis
asked Nov 18 at 22:12
Mike
2,624211
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to prove or disprove the following:
Let $G$ be a non-bipartite AND connected $k$-regular graph where the size of the second-largest eigenvalue of the adjacency matrix $A_G$ of $G$ is $lambda$ (which as $G$ is connected is always strictly less than $k$). Then every eigenvalue of $A_G$, except for one, has absolute value no larger than $|lambda|$.
Of course, the shaded above would not be true if $G$ is allowed to be bipartite; if $G$ is $k$-regular bipartite the eigenvalues of $A_G$ are symmetric around 0, so $-k$ is an eigenvalue of $A_G$ for any $k$-regular bipartite $G$. I am wondering if there is a $k$-regular connected graph $G$ that is not bipartite that has smallest eigenvalue in the interval $(-k,-|lambda|]$, where $lambda$ is as above the 2nd-largest eigenvalue of $A_G$.
I have been looking around the internet for a solution either way, meanwhile my matrix analysis is rusty and I know we have some really smart people here. Anyone have any insight into this?
Thanks!
combinatorics graph-theory matrix-analysis
asked Nov 18 at 22:12
Mike
2,624211
I am trying to prove or disprove the following:
Let $G$ be a non-bipartite AND connected $k$-regular graph where the size of the second-largest eigenvalue of the adjacency matrix $A_G$ of $G$ is $lambda$ (which as $G$ is connected is always strictly less than $k$). Then every eigenvalue of $A_G$, except for one, has absolute value no larger than $|lambda|$.
Of course, the shaded above would not be true if $G$ is allowed to be bipartite; if $G$ is $k$-regular bipartite the eigenvalues of $A_G$ are symmetric around 0, so $-k$ is an eigenvalue of $A_G$ for any $k$-regular bipartite $G$. I am wondering if there is a $k$-regular connected graph $G$ that is not bipartite that has smallest eigenvalue in the interval $(-k,-|lambda|]$, where $lambda$ is as above the 2nd-largest eigenvalue of $A_G$.
I have been looking around the internet for a solution either way, meanwhile my matrix analysis is rusty and I know we have some really smart people here. Anyone have any insight into this?
Thanks!
combinatorics graph-theory matrix-analysis
combinatorics graph-theory matrix-analysis
asked Nov 18 at 22:12
Mike
2,624211
asked Nov 18 at 22:12
Mike
2,624211
edited Nov 18 at 22:17
asked Nov 18 at 22:12
Mike
2,624211
asked Nov 18 at 22:12
Mike
2,624211
asked Nov 18 at 22:12
Mike
2,624211
2,624211
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1 Answer
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The eigenvalues of the Petersen graph are $3$, $1$ and $-2$.
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
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up vote
1
down vote
The eigenvalues of the Petersen graph are $3$, $1$ and $-2$.
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
add a comment |
up vote
1
down vote
The eigenvalues of the Petersen graph are $3$, $1$ and $-2$.
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
add a comment |
up vote
1
down vote
up vote
1
down vote
The eigenvalues of the Petersen graph are $3$, $1$ and $-2$.
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
The eigenvalues of the Petersen graph are $3$, $1$ and $-2$.
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
answered Nov 19 at 1:34
Chris Godsil
11.5k21634
11.5k21634
add a comment |
add a comment |
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