# If there are eigenvectors with largest components $i$ resp. $j$, then is there an eigenvector with two largest components $i$ and $j$?

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 $$G$$.

• @dodd Correct me if I am wrong, but I think this follows from the Theorem of Perron-Frobenius, at least for connected graphs. If it makes a difference, I should restrict my question to connected graphs, but I do not think so. Jan 5 at 2:02
• Try the graph with two vertices and no edges. It has one eigenvalue of multiplicity 2. Jan 5 at 2:16
• @dodd I restricted the question to connected graphs now. Jan 5 at 2:18
• @dodd This subspace is not just any subspace, but it is a very special eigenspace of an irreducible symmetric 01-matrix, and I ask for the existence of $u_{ij}$ only if the $(i,j)$-entry of that matrix is one. I consider the formulation in terms of graphs more natural than just talking about this matrix (and I do not see how this can be reasonably formulated just in terms of subspaces). Of course, I am happy with an answer in any language, whether graphs, matrices, subspaces, etc. Jan 5 at 2:38
• @dodd For an arbitrary subspace there is no meaning in "an edge $ij\in E$". But you could consider the smallest eigenvalue (that is, the corresponding eigenspace) of the 5-cycle graph. Then the $u_i$ exist for all vertices, but $u_{ij}$ exists for no edge. Jan 5 at 10:30

$$\quad\quad$$
It turns out that this exact figure depicts the spectral embedding of this graph to the second-largest eigenvalue $$\theta_2$$, and we see that some of its edges are not embedded in edges of the rsulting polytope, but merely in its faces.
So, for example, if $$ij\in E$$ are choose note as an edge of the dodecahedron, but as an edge of a pentagram, then the $$\smash{u_i,u_j\in\mathrm{Eig}_G(\theta_2)}$$ exist, but $$u_{ij}$$ does not.