## Return to Answer

4 deleted 1 characters in body

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 and 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, $\mu_3(K_3)=\mu_3(P_3)=3$.

3 An attempt to make the text properly formatted

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 and eigenvector $x'$ of $G'$, belonging to the eigenvalue $\mu_n(G')$. Indexing coordinates by the vertices of $G$, we have

$$\mu_n(G') = \frac{\sum_{(u,v)\in E'} (x'_u-x'v)^2}{\|x'\|^2} x'_u-x'_v)^2}{\|x'\|^2} \le \frac{\sum{(u,v)\in frac{\sum_{(u,v)\in E} (x'_u-x'v)^2}{\|x'\|^2} x'_u-x'_v)^2}{\|x'\|^2} \le \sup{x\ne sup_{x\ne 0} \frac{\sum_{(u,v)\in E} (x_u-x_v)^2}{\|x\|^2} = \mu_n(G).$$

Notice that the inequality may fail to be strict in the Laplacian case; say, $\mu_3(K_3)=\mu_3(P_3)=3$.

2 added 838 characters in body; deleted 2 characters in body

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 and 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, $\mu_3(K_3)=\mu_3(P_3)=3$.

1