I came across a Chinese reference in the paper "On the spectral radius of trees with ﬁxed diameter" by Guo and Shao. The attribute the following to Q. Li, K.Q. Feng in: "On the largest eigenvalue of graphs", Acta Math. Appl. Sinica 2 (1979) 167–175. 


Let $\lambda\_1(H)$ be the spectral radius of the adjacency matrix of the graph H.


Let $G$ be a connected graph, and let $G'$ be a proper subgraph of $G$. Then $\lambda\_1(G)>\lambda\_1(G')$. 


1) Is there a trivial argument of this using interlacing or is something more sophisticated needed here.


2) Is this also true for the Laplacian spectral radius. I assume if there is an interlacing argument this will be trivial.



Spectral radius of a proper subgraph
Here is a simple proof. Without loss of generality, $G'$ is obtained from $G$
by deleting some edges (and keeping all vertices). Let $A$ and $A'$ denote
the adjacency matrices of $G$ and $G'$, respectively, and let $x'$ be an
eigenvector of $A'$, belonging to the eigenvalue $\lambda\_1(G')$, such that
all coordinates of $x'$ are non-negative. We have then
 $$ \lambda\_1(G') = \frac{\langle x',A'x'\rangle}{\|x'\|^2}
 \le \frac{\langle x',Ax'\rangle}{\|x'\|^2}
 \le \sup\_{x\ne 0} \frac{\langle x,Ax\rangle}{\|x\|^2}
 = \lambda\_1(G); $$
indeed, if all coordinates of $x'$ are strictly positive, then the first
inequality is strict, and if $x'$ has zero coordinates, then the second
inequality is strict (by Perron-Frobenius, which says that the supremum is
attained on a vector with all coordinates distinct from $0$).


The largest *Laplacian* eigenvalue (which, of course, is equal to the Laplacian spectral radius) can be dealt with in a similar manner. Suppose that $G'$ is obtained from $G$ by deleting some edges, and let $E$ and $E'$ be the edge sets of $G$ and $G'$, respectively. Furthermore, denote by $\mu\_n(G)$ and $\mu\_n(G')$ the largest Laplacian eigenvalues of $G$ and $G'$, and fix an eigenvector $x'$ of $G'$, belonging to the eigenvalue $\mu\_n(G')$. Indexing coordinates by the vertices of $G$, we have 


$$ \mu\_n(G') = \frac{\sum\_{(u,v)\in E'} (x'\_u-x'\_v)^2}{\|x'\|^2}
 \le \frac{\sum\_{(u,v)\in E} (x'\_u-x'\_v)^2}{\|x'\|^2} 
 \le \sup\_{x\ne 0} \frac{\sum\_{(u,v)\in E} (x\_u-x\_v)^2}{\|x\|^2}
 = \mu\_n(G). $$


Notice that the inequality may fail to be strict in the Laplacian case; say, $\mu\_3(K\_3)=\mu\_3(P\_3)=3$.


