Hadamard $\ell_2$ sum of two symmetric positive semidefinite matrices

Question

This is a follow-up question to this and this.

Let $A=(a_{ij})$ and $B=(b_{ij})$ be symmetric positive semidefinite $n\times n$ matrices such that all $a_{ij}\geq 0$, $b_{ij}\geq 0$ and $a_{ii}=b_{ii}=1$. Let $C=((a_{ij}^2+b_{ij}^2)^{1/2})$ and let $\lambda_n(C)$ be the smallest eigenvalue of $C$. Let $$\Lambda_n:=\min_{A,B}\lambda_n(C).$$ Is $\lim_{n\to\infty}\Lambda_n=-\infty$?

Edit: For any $s,t\geq 0$ and $x,y>0$, the example below with $$C=\Bigl(\sqrt{(s+x\cdot a_{ij})^2+(t+y\cdot b_{ij})^2}\Bigr)$$ also satisfies $\lim_{n\to\infty}\lambda_n(C)=-\infty$.

Answer 1

The answer is yes. If $I$ is an $n\times n$ matrix with all entries equal to $1$ and $$ A=\left[\begin{array}{cccc} I & I & 0 & 0 \\ I & I & 0 & 0 \\ 0 & 0 & I & I \\ 0 & 0 & I & I \end{array}\right],\,\, B=\left[\begin{array}{cccc} I & 0 & I & 0 \\ 0 & I & 0 & I \\ I & 0 & I & 0 \\ 0 & I & 0 & I \end{array}\right] $$ then the smallest eigenvalue of $$ C=\left[\begin{array}{cccc} \sqrt{2}I & I & I & 0 \\ I & \sqrt{2}I & 0 & I \\ I & 0 & \sqrt{2}I & I \\ 0 & I & I & \sqrt{2}I \end{array}\right] $$ is less than or equal to $-(2-\sqrt{2})n$. This can be seen by computing $(Cx,x)$ with $x=(e,-e,-e,e)^T$, where $e$ is a $1\times n$ vector with all entries equal to $1/\sqrt{4n}$. This is an extension of the example of fedja in this question.