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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:

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

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The matrices are of the form $$ A=\begin{pmatrix} 1 & C \\ C^* & D \end{pmatrix}, \quad\quad B= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} A \begin{pmatrix} 1 & 0\\ 0&2\end{pmatrix} , $$ with the blocks corresponding to the sizes of the sets $M_j$ involved.

Let $v=(x,y)^t$ be a normalized eigenvector for the minimum eigenvalue $\lambda $ of $B$. Then $$ (x,2y)A\begin{pmatrix} x \\ 2y \end{pmatrix} = \lambda $$ also, but this modified vector has larger norm.

So the desired inequality $\lambda_j(A)\le\lambda_j(B)$ (for all eigenvalues, not just the first one) will follow if we can show that $B\ge 0$. This is true for $y=1$ because in this case we can interpret $$ |M_j\cap M_k|=\sum_n \chi_j(n)\chi_k(n) $$ as the scalar product in $\ell^2$ of the characteristic functions, and this makes $v^*Bv$ equal to $\|f\|_2^2\ge 0$, with $f=\sum v_j\chi_j$.

For general $y>0$, we have $B(y)=(1/y)B(1) + D$, for a diagonal matrix $D$ with non-negative entries.

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Claim. $\lambda_\min(A_N) \le 4\lambda_\min(B_N)$.

Proof. Let $C_N:=\bigl[\tfrac{1}{|M_i||M_j|}\bigr]$. Then, $B_N = A_N \circ C_N$, where $\circ$ denotes the Hadamard product. Observe that by construction both $A_N$ and $C_N$ are positive semidefinite, so $B_N$ is also psd. Let's drop the subscript $N$ for brevity. Define $c = \text{diag}(C)$ sorted in decreasing order, so in particular, $c_\min = \min_{1\le i \le N} 1/|M_i|^2=1/4$.

Now, from Theorem 3(ii) of Bapat and Sunder, it follows that: \begin{equation*} \lambda_\min(B)=\lambda_\min(A \circ C) \ge \lambda_\min(A)c_\min = \lambda_\min(A_N)/4. \end{equation*}


Note: The result of Bapat and Sunder is more general. For psd matrices $A$ and $C$ it states that \begin{equation*} \prod_{j=k}^n \lambda_j(A\circ C) \ge \prod_{j=k}^n\lambda_j(A)c_j, \end{equation*} where $1\le k \le n$, and $\lambda_1(\cdot)\ge \lambda_2(\cdot) \ge \cdots \ge \lambda_n(\cdot)$, while $c$ is as above.

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  • $\begingroup$ Also: do you mean positive semi definite or entrywise positive semi definite? $\endgroup$ – lcv Oct 24 '18 at 7:46
  • $\begingroup$ @ChristianRemling Well, each entry of $A$ and $C$ can be written as an inner product (as you now also are doing in your answer)....hence the semi-definiteness $\endgroup$ – Suvrit Oct 24 '18 at 10:11
  • $\begingroup$ @lcv I mean the usual notation of psd-ness. $\endgroup$ – Suvrit Oct 24 '18 at 10:12
  • $\begingroup$ @ChristianRemling sorry, my bad; somehow I thought that because it is intersection, it's obvious for $A$, and of course $C=zz^T$, so that's also clear. $\endgroup$ – Suvrit Oct 24 '18 at 19:04

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