This is a follow-up question to [this](https://mathoverflow.net/questions/335996/hadamard-ell-p-sum-of-two-symmetric-positive-semidefinite-matrices) and [this](https://mathoverflow.net/questions/336099/hadamard-ell-p-sum-of-two-symmetric-positive-semidefinite-matrices-follow-up).


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 $\lim\_{n\to\infty}\lambda\_n(C)=-\infty$.



Hadamard $\ell_2$ sum of two symmetric positive semidefinite matrices
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](https://mathoverflow.net/questions/336099/hadamard-ell-p-sum-of-two-symmetric-positive-semidefinite-matrices-follow-up?noredirect=1&lq=1) question.


