suppose that the difference of edges of two graph $G$ and $G^{'}$ is 1,show that $|\lambda_{max}(G)-\lambda_{max}(G^{'})|\leq1$.
$\lambda_{max}$ is the biggest eigenvalue of adjacency matrix of corresponding graph.
any help or hint will be great,thanks.
my Idea :because $\lambda_{max} \leq \Delta$ ,if we omit on edge from $G$ ,maximum degree will be same or reduce by one,if it will be same so the difference is zero so their difference is fewer than 1,and in other case we have $\lambda_{max}(G) - \lambda_{max}(G^{'}) \leq \Delta(G)-(\Delta(G)-1)=1$
The star $K_{1,n}$ has spectral radius $\sqrt{n}$, while the path on $n$ vertices has spectral radius less than 2. So your claim fails when $G$ is the star on $n+1$ vertices and $G'$ is the path on $n+2$ vertices and $n\ge9$. (And for all larger values of $n$.