The maximal \alpha-index of trees with k pendent vertices and its computation

Let $G$ be a graph with adjacency matrix $A(G)$ and let $D(G)$ be the diagonal matrix of the degrees of $G$. The $\alpha-$ index of $G$ is the spectral radius $\rho_{\alpha}\left( G\right)$ of the matrix $A_{\alpha}\left( G\right)=\alpha D\left( G\right) +(1-\alpha)A\left( G\right)$ where $\alpha \in [0,1]$. Let $T_{n,k}$ be the tree of order $n$ and $k$ pendent vertices obtained from a star $K_{1,k}$ and $k$ pendent paths of almost equal lengths attached to different pendent vertices of $K_{1,k}$. It is shown that if $\alpha\in\left[ 0,1\right) $ and $T$ is a tree of order $n$ with $k$ pendent vertices then% \[ \rho_{\alpha}(T)\leq\rho_{\alpha}(T_{n,k}), \] with equality holding if and only if $T=T_{n,k}$. This result generalizes a theorem of Wu, Xiao and Hong \cite{WXH05} in which the result is proved for the adjacency matrix ($\alpha=0$). Let $q=[\frac{n-1}{k}]$ and $n-1=kq+r$, $0 \leq r \leq k-1$. It is also obtained that the spectrum of $A_{\alpha}(T_{n,k})$ is the union of the spectra of two special symmetric tridiagonal matrices of order $q$ and $q+1$ when $r=0$ or the union of the spectra of three special symmetric tridiagonal matrices of order $q$, $q+1$ and $2q+2$ when $r \neq 0$. Thus the $\alpha-$ index of $T_{n,k}$ can be computed as the largest eigenvalue of the special symmetric tridiagonal matrix of order $q+1$ if $r=0$ or order $2q+2$ if $r\neq 0$.

A Test Matrix for an Inverse Eigenvalue Problem

We present a real symmetric tridiagonal matrix of ordernwhose eigenvalues are{2k}k=0n-1which also satisfies the additional condition that its leading principle submatrix has a uniformly interlaced spectrum,{2l+1}l=0n-2. The matrix entries are explicit functions of the sizen, and so the matrix can be used as a test matrix for eigenproblems, both forward and inverse. An explicit solution of a spring-mass inverse problem incorporating the test matrix is provided.