Let $G=(V,E)$ be a (finite simple) graph with vertex set $V=\{1,...,n\}$ and let $\theta_2\in\Bbb R$ be the second-largest eigenvalue of its adjacency matrix. I wonder about the following question:

Fix an edge $ij\in E$.

Suppose that there are $\theta_2$-eigenvectors $u_i,u_j\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ so that the largest component of $u_i$ is the $i$-th component, and the largest component of $u_j$ is the $j$-th component.

Question: Is there an eigenvector $u_{ij}\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ with exactly two identical largest components, namely, the $i$-th component and the $j$-th component?

I think (but I do not know) that if it is possible at all, then one can choose $u_{ij}=\alpha u_i + \beta u_j$ as a convex combination.

If you are familiar with the term eigenpolytope, then this can be formulated as follows: if $i$ and $j$ are embedded as vertices of the $\theta_2$-eigenpolytope, then is $ij\in E$ embedded as an edge of the eigenpolytope?

The choice of the second-largest eigenvalue is important: it is not generally true for any other eigenvalue (except, trivially, for the largest eigenvalue $\theta_1$, or any other eigenvalue of multiplicity 1). In contrast, I have not found a single counterexample for $\theta_2$. It has been proven in special cases, e.g. for distance-regular graphs. It is easy to construct counter-examples if one allows edge-weights on $G$.

Let $G=(V,E)$ be a connected (finite simple) graph with vertex set $V=\{1,...,n\}$ and let $\theta_2\in\Bbb R$ be the second-largest eigenvalue of its adjacency matrix. I wonder about the following question:

Fix an edge $ij\in E$.

Suppose that there are $\theta_2$-eigenvectors $u_i,u_j\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ so that the largest component of $u_i$ is the $i$-th component, and the largest component of $u_j$ is the $j$-th component.

Question: Is there an eigenvector $u_{ij}\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ with exactly two identical largest components, namely, the $i$-th component and the $j$-th component?

I think (but I do not know) that if it is possible at all, then one can choose $u_{ij}=\alpha u_i + \beta u_j$ as a convex combination.

If you are familiar with the term eigenpolytope, then this can be formulated as follows: if $i$ and $j$ are embedded as vertices of the $\theta_2$-eigenpolytope, then is $ij\in E$ embedded as an edge of the eigenpolytope?

The choice of the second-largest eigenvalue is important: it is not generally true for any other eigenvalue (except, trivially, for the largest eigenvalue $\theta_1$, or any other eigenvalue of multiplicity 1). In contrast, I have not found a single counterexample for $\theta_2$. It has been proven in special cases, e.g. for distance-regular graphs. It is easy to construct counter-examples if one allows edge-weights on $G$.

Let $G=(V,E)$ be a (finite simple) graph with vertex set $V=\{1,...,n\}$ and let $\theta_2\in\Bbb R$ be the second-largest eigenvalue of its adjacency matrix. I wonder about the following question:

Fix an edge $ij\in E$.

Suppose that there are $\theta_2$-eigenvectors $u_i,u_j\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ so that the largest component of $u_i$ is the $i$-th component, and the largest component of $u_j$ is the $j$-th component.

Question: Is there an eigenvector $u_{ij}\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ with exactly two identical largest components, namely, the $i$-th component and the $j$-th component?

I think (but I do not know) that if it is possible at all, then one can choose $u_{ij}=\alpha u_i + \beta u_j$ as a convex combination.

If you are familiar with the term eigenpolytope, then this can be formulated as follows: if $i$ and $j$ are embedded as vertices of the $\theta_2$-eigenpolytope, then is $ij\in E$ embedded as an edge of the eigenpolytope?

The choice of the second-largest eigenvalue is important: it is not generally true for any other eigenvalue. In contrast, I have not found a single counterexample for $\theta_2$. It has been proven in special cases, e.g. for distance-regular graphs. It is easy to construct counter-examples if one allows edge-weights on $G$.

Let $G=(V,E)$ be a (finite simple) graph with vertex set $V=\{1,...,n\}$ and let $\theta_2\in\Bbb R$ be the second-largest eigenvalue of its adjacency matrix. I wonder about the following question:

Fix an edge $ij\in E$.

Suppose that there are $\theta_2$-eigenvectors $u_i,u_j\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ so that the largest component of $u_i$ is the $i$-th component, and the largest component of $u_j$ is the $j$-th component.

Question: Is there an eigenvector $u_{ij}\in\mathrm{Eig}_G(\theta_2)\subseteq\Bbb R^n$ with exactly two identical largest components, namely, the $i$-th component and the $j$-th component?

I think (but I do not know) that if it is possible at all, then one can choose $u_{ij}=\alpha u_i + \beta u_j$ as a convex combination.

If you are familiar with the term eigenpolytope, then this can be formulated as follows: if $i$ and $j$ are embedded as vertices of the $\theta_2$-eigenpolytope, then is $ij\in E$ embedded as an edge of the eigenpolytope?

The choice of the second-largest eigenvalue is important: it is not generally true for any other eigenvalue (except, trivially, for the largest eigenvalue $\theta_1$, or any other eigenvalue of multiplicity 1). In contrast, I have not found a single counterexample for $\theta_2$. It has been proven in special cases, e.g. for distance-regular graphs. It is easy to construct counter-examples if one allows edge-weights on $G$.