Let G be a connected simple graph with the vertex set $V\left(G\right)=\{v_1,v_2,\dots ,v_n\}$, where the distance $d_G(v_i,v_j)$ is the length of a shortest path between $v_i$ and $v_j$, and the distance matrix $D\left(G\right)=\left(d_{ij}\right)n\times n$ is defined by $d_{ij}=d_G(v_i,v_j)$. Being real symmetric and non-negative, $D\left(G\right)$ has real eigenvalues ${\lambda }_1\left(G\right)\ge {\lambda }_2\left(G\right)\ge \dots \ge {\lambda }_n\left(G\right)$, with ${\lambda }_n\left(G\right)$ referred to as the least distance eigenvalue of G. In this paper, we give the maximum value of the least distance eigenvalue, and determine the unique extremal graph with two pendant vertices whose complements attain the value.

Suppose $G$ is a connected simple graph with the vertex set $V( G ) = \{ v_1,v_2,\cdots ,v_n \} $. Let $d_G( v_i,v_j ) $ be the least distance between $v_i$ and $v_j$ in $G$. Then the distance matrix of $G$ is $D( G ) =( d_{ij} ) _{n\times n}$, where $d_{ij}=d_G( v_i,v_j ) $. Since $D( G )$ is a non-negative real symmetric matrix, its eigenvalues can be arranged as $λ_1(G)\ge λ_2(G)\ge \cdots \ge λ_n(G)$, where eigenvalue $λ_n(G)$ is called the least distance eigenvalue of $G$. In this paper we determine the unique graph whose least distance eigenvalue attains maximum among all complements of graphs of diameter greater than three.