On trees with exactly one characteristic element (Q869893): Difference between revisions

\loadeusb For a given tree \(T\), its Laplacian matrix \(L\) and a given vector \(Y\) (a valuation of vertices of \(T\)), a vertex \(u\) of \(T\) is called a characteristic vertex if and only if \(Y[u]=0\) and there is a vertex \(w\) adjacent to \(u\) with \(Y[w]\neq 0\); and an edge \(\{u,w\}\) is called a characteristic edge if \(Y[u]Y[w]<0\). The characteristic set of \(T\) with respect to \(Y\), \({\mathcal S}(T,Y)\), is a collection of all characteristic vertices and characteristic edges of \(T\). Let \(\lambda_k\) be the \(k\)th smallest eigenvalue. If \(\lambda_k>\lambda_{k-1}\), the corresponding eigenvector \(Y\) of \(\lambda_k\) is called a \(k\)-vector and a tree \(T\) is called \(k\)- simple if and only if \(| {\mathcal S}(T,Y)| =1\) for all \(k\)-vectors \(Y\). The authors show the existence of \(k\)-simple trees and characterize them. They also show that the characteristic sets determined by all the \(k\)-vectors are the same and give some properties of the eigenvalues and eigenvectors of a \(k\)-simple tree.