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M. Winter
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When do antipodal Antipodal vertices in a graph have metrically opposite spectral graph embeddings?

Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following holdshold:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$

  • is walk-regular, (resp. vertex-transitive)
  • is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive),
  • is distance-regular (resp. distance-transitive), or
  • has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.

Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$). But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.

When do antipodal vertices in a graph have metrically opposite spectral embeddings?

Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following holds:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$

  • is walk-regular, (resp. vertex-transitive)
  • is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive),
  • is distance-regular (resp. distance-transitive), or
  • has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.

Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$). But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.

Antipodal vertices in spectral graph embeddings

Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following hold:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$

  • is walk-regular, (resp. vertex-transitive)
  • is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive),
  • is distance-regular (resp. distance-transitive), or
  • has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.

Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$). But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.

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Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70

Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following holds:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$ is:

  • is walk-regular, (resp. vertex-transitive)
  • is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive), or
  • is distance-regular (resp. distance-transitive), or
  • has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.

Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$). But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.

Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following holds:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$ is:

  • walk-regular, (resp. vertex-transitive)
  • 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive), or
  • distance-regular (resp. distance-transitive).

Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following holds:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$

  • is walk-regular, (resp. vertex-transitive)
  • is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive),
  • is distance-regular (resp. distance-transitive), or
  • has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.

Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$). But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.

Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70

When do antipodal vertices in a graph have metrically opposite spectral embeddings?

Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following holds:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$ is:

  • walk-regular, (resp. vertex-transitive)
  • 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive), or
  • distance-regular (resp. distance-transitive).