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Abstract:It is well known that the spectral radius $\rho(T)$ of a tree $T$ with at least $3$ vertices has the property that $\frac 14\rho(T)^2+1<\Delta(T)\le \rho(T)^2$, where $\Delta(T)$ is the maximum degree of $T$. Let $\mathbb{P}$ denote the set of spectral radii of all non-trivial trees. In this article, we study the inverse problem that for any $\alpha\in \mathbb{P}$ and integer $r$ satisfying the condition $\frac 14\alpha^2+1<r\le \alpha^2$, is there a tree $T$ such that $\Delta(T)=r$ and $\rho(T)=\alpha$?
For any positive integer $r$ and positive number $\alpha$, let ${\mathscr W}_r(\alpha)$ denote a set of non-negative real numbers defined as follows: $\alpha\in {\mathscr W}_r(\alpha)$, and for any multi-set $\{q_i\in {\mathscr W}_r(\alpha): q_i>0, 1\le i\le s\}$, if $s<r$ and $\beta:=\alpha-\sum\limits_{i=1}^sq_i^{-1}\ge 0$, or $s=r$ and $\beta=0$, then $\beta \in {\mathscr W}_r(\alpha)$. We first show that $0\in {\mathscr W}_r(\alpha)$ if and only if there exists a tree $T$ with $\Delta(T)\le r$ and $\rho(T)=\alpha$. It follows directly that $\mathbb{P}$ is exactly the set of positive numbers $\alpha$ such that $0\in {\mathscr W}_{\lfloor\alpha^2\rfloor}(\alpha)$. Applying this conclusion, we prove that for any two positive integers $r\ge 2$ and $k$, there exists a tree $T$ with $\Delta(T)=r$ and $\rho(T)=\sqrt k$ if and only if $\frac 14 k+1<r\le k$.

Submission history

From: Fengming Dong [view email]
[v1] Wed, 9 Apr 2025 06:30:32 UTC (18 KB)
[v2] Mon, 21 Jul 2025 07:57:48 UTC (22 KB)
[v3] Fri, 12 Sep 2025 08:22:35 UTC (17 KB)
[v4] Mon, 24 Aug 2026 05:42:28 UTC (20 KB)