More edges implies larger eigenvalues of graph Laplacian?

Question

Suppose

1. $G_1$ and $G_2$ are two connected graphs with the same nodes.
2. $G_2 \subset G_1$. $G_2$ is the connected subgraph of $G_1$.

I know that $\lambda_2(L(G_2))\leq \lambda_2(L(G_1))$ since $\lambda_2$ represents the connectivity. ($L$ represents the graph Laplacian matrix.)

Can we say the following

1. $\lambda_i (L(G_2))\leq \lambda_i(G_1)$ for all $i$?
2. $L(G_1) \succeq L(G_2)$? (We know graph Laplacian matrix is positive semi-definite.)

I believe 1. implies 2.

If you know any reference of above, please let me know.

Thanks in advance.

Note: By my old and not clear memory, I guess this is true in graph theory. But I forgot where I read it

Answer 1

This is a standard interlacing problem. We will use the well-known interlacing theorem (it is a specific case of Poincaré separation theorem):

If $B$ is a principal submatrix of a symmetric matrix $A$, then the eigenvalues of $B$ interlace the eigenvalues of $A$.

Let $G$ be a graph on $n$ vertices and let $H$ be a spanning subgraph (this is your case as $G_1$ and $G_2$ have same vertex set). I claim that for $i\in\{1,\ldots,n\}$, $\lambda_i(L(H)) \leq \lambda(L(G))$.

Proof : Let $N$ be the directed incidence matrix obtained by a random orientation the edges of $G$. Then it is easy to verify that $L = NN^\perp$, and that $NN^\perp$ and $N^\perp N$ have the same eigenvalues. Removing an edge from $G$ results in removing a column from $N$ and yields a principal submatrix of $N^\perp N$. Therefore we can apply Cauchy's interlacing theorem and the result holds. $\square$