Let $H$ be a graph and let $G=H \vee K\_{1}$ be obtained by creating a new vertex and joining it to every vertex in $H$. 


This situation has many different names: $G$ is called the *cone* or the *suspension* of $H$; the new vertex is called *dominating* or *universal* - and probably there are other names that have slipped my mind right now.


My question concerns the difference between the spectral (adjacency) radii of $G$ and $H$. Clearly $\rho(G) > \rho(H)$. Now I vaguely recall seeing somewhere the inequality



> 
> $$ \rho(G)-\rho(H) \geq 1. $$
> 
> 
> 


But I am unable to track it down. Is it true? And if yes, is it known?



The spectral radius of a modified graph
Yes, this is true, but I don't know a reference, so here's a proof (I think). Let
$$
R(A, x) = \frac{x^T A x}{x^T x}
$$
be the Rayleigh quotient. We know that for a symmetric (in fact Hermitian) matrix $A$ and any vector $x$,
$$
R(A, x) \leq \rho(A)
$$
with equality if and only if $x$ is the Perron vector for $A$.


Now suppose that $H$ is your original graph, and that $G = H \vee K\_1$ is the cone, and that $A(H)$, $A(G)$ are their respective adjacency matrices. Then let $\lambda = \rho(H)$ and suppose that $v$ is a vector such that $A(H) v = \lambda v$, and recall that all the entries of $v$ can be taken to be non-negative. Renumbering if necessary we can assume
that 
$$
A(G) = \left[
\begin{array}{cc}
A(H)&j\\
j^T&0
\end{array}
\right],
$$
where $j$ is the all-ones vector.


Our aim is to exhibit a vector $w$ such that $R(A(G), w) \geq \lambda+1$ thus proving the inequality. So let $w$ be the vector obtained by adjoining an additional coordinate equal to $1$ to $v$. It then follows that
$$
A(G) w= \left[
\begin{array}{cc}
A(H)&j\\
j^T&0
\end{array}
\right]
\left[
\begin{array}{c}
v\\
1
\end{array}
\right]
= 
\left[
\begin{array}{c}
\lambda v + j\\
j^T v
\end{array}
\right]
$$
and 
$$
w^T w = v^T v + 1.
$$
Therefore the Rayleigh quotient
$$
R(A(G), w) = \frac{\lambda v^T v + 2 j^T v} { v^T v + 1}.
$$


Now assume that $v$ is normalised so that its maximum entry is $1$ and again recall that every entry of $v$ is non-negative. Then we have
the inequalities $j^T v \geq v^T v$ (because each $0 \leq v\_i \leq 1$) and $j^T v \geq (\lambda +1)$ (because if $i$ is the index such that $v\_i = 1$ then $(Av)\_i = \lambda$ and $j^T v \geq (Av)\_i + v\_i$). Putting it all together we have
$$
R(A(G), w) \geq \frac{\lambda v^T v + v^T v + (\lambda + 1)}{v^Tv + 1} = \lambda+1.
$$


