Consider the set $\mathbf{N}:=\left\{1,2,....,N \right\}$ and let $$\mathbf M:=\left\{ M_i; M_i \subset \mathbf N \text{ such that } \left\lvert M_i \right\rvert=2 \text{ or }\left\lvert M_i \right\rvert=1 \right\}$$ be the set of all subsets of $\mathbf{N}$ that are of cardinality $1$ or $2.$

The cardinality of the set $\mathbf M$ itself is $\binom{n}{1}+\binom{n}{2}=:K$

We can then study for $y \in (0,1)$ the $K \times K$ matrix

$$A_N = \left( \frac{\left\lvert M_i \cap M_j \right\rvert}{\left\lvert M_i \right\rvert\left\lvert M_j \right\rvert}y^{-\left\lvert M_i \cap M_j \right\rvert} \right)_{i,j}$$

and

$$B_N = \left( \left\lvert M_i \cap M_j \right\rvert y^{-\left\lvert M_i \cap M_j \right\rvert} \right)_{i,j}.$$

**Question**
I conjecture that $\lambda_{\text{min}}(A_N)\le \lambda_{\text{min}}(B_N)$ for any $N$ and would like to know if one can actually show this?

As a first step, I would like to know if one can show that $$\lambda_{\text{min}}(A_N)\le C\lambda_{\text{min}}(B_N)$$ for some $C$ independent of $N$?

In fact, I am **not** claiming that $A_N \le B_N$ in the sense of matrices. But it seems as if the eigenvalues of $B_N$ are shifted up when compared with $A_N.$

**Numerical evidence:**

For $N=2$ we can explicitly write down the matrices $$A_2 =\left( \begin{array}{ccc} \frac{1}{y} & 0 & \frac{1}{2 y} \\ 0 & \frac{1}{y} & \frac{1}{2 y} \\ \frac{1}{2 y} & \frac{1}{2 y} & \frac{1}{2 y^2} \\ \end{array} \right) \text{ and }B_2 = \left( \begin{array}{ccc} \frac{1}{y} & 0 & \frac{1}{y} \\ 0 & \frac{1}{y} & \frac{1}{y} \\ \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} \\ \end{array} \right)$$

We obtain for the lowest eigenvalue of $A_2$ (orange) and $B_2$(blue) as a function of $y$

For $N=3$ we get qualitatively the same picture, i.e. the lowest eigenvalue of $A_3$ remains below the lowest one of $B_3$:

In this case:

$$A_3=\left( \begin{array}{cccccc} \frac{1}{y} & 0 & 0 & \frac{1}{2 y} & 0 & \frac{1}{2 y} \\ 0 & \frac{1}{y} & 0 & \frac{1}{2 y} & \frac{1}{2 y} & 0 \\ 0 & 0 & \frac{1}{y} & 0 & \frac{1}{2 y} & \frac{1}{2 y} \\ \frac{1}{2 y} & \frac{1}{2 y} & 0 & \frac{1}{2 y^2} & \frac{1}{4 y} & \frac{1}{4 y} \\ 0 & \frac{1}{2 y} & \frac{1}{2 y} & \frac{1}{4 y} & \frac{1}{2 y^2} & \frac{1}{4 y} \\ \frac{1}{2 y} & 0 & \frac{1}{2 y} & \frac{1}{4 y} & \frac{1}{4 y} & \frac{1}{2 y^2} \\ \end{array} \right)\text{ and } B_3=\left( \begin{array}{cccccc} \frac{1}{y} & 0 & 0 & \frac{1}{y} & 0 & \frac{1}{y} \\ 0 & \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} \\ \frac{1}{y} & \frac{1}{y} & 0 & \frac{2}{y^2} & \frac{1}{y} & \frac{1}{y} \\ 0 & \frac{1}{y} & \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} & \frac{1}{y} \\ \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} \\ \end{array} \right)$$