For symmetric real matrices, each $L^2$ circle contains at least an eigenvalue? Let $A=(a_{ij})_{ij}$ be a symmetric real matrix. I know by the Gershgorin circle theorem, every eigenvalue $\lambda$ of $A$ must lie in a closed circle defined as
$$D_1(A,i):=\left\{x\in\mathbb{R}\left||x-a_{ii}|\le\sum_{j\neq i}|a_{ij}|\right.\right\}$$
for some $i$. However, the converse does not necessarily hold, which means for each index $i$, it is possible that $D_1(A,i)$ does not contain any eigenvalue of $A$.
Here is my question, if we choose the $L^2$ norm rather than the $L^1$ norm to define the circles
$$D_2(A,i):=\left\{x\in\mathbb{R}\left||x-a_{ii}|^2\le\sum_{j\neq i}|a_{ij}|^2\right.\right\},$$
then for each index $i$, $D_2(A,i)$ must contain at least one eigenvalue of $A$. I have tried to search for it online, and it turns out a more general version of the conclusion can be found in this paper (by setting $x_i=y_i=\left(\sum_{j\neq i}|a_{ij}|^2\right)^{1/2}$ in Lemma 3.1). Of course I can follow the proof of the Lemma directly, but I am wondering if there exists a simpler and easier approach since the proposition itself is not complex at all.
Any help would be appreciated.
 A: A word of caution: the lemma of Barnes and Hoffman is not a generalisation of Gerschgorin disc theorem, because the union of all Gerschgorin discs always covers the spectrum of a matrix, but this is not the case in Barnes and Hoffman's lemma. E.g. when $A$ is the $3\times3$ matrix of ones and $x_i=y_i=\left(\sum_{j\ne i}|a_{ij}|^2\right)^{1/2}$, the eigenvalue $3$ lies outside $D_2(A,i)$ for all $i$.
However, since the Euclidean norm is bounded above by the $\ell_1$ norm, the lemma of Barnes and Hoffman implies that each Gerschgorin disc must contain some eigenvalue of $A$ when $A$ is real symmetric. (This in general is false when $A$ is not Hermitian.)
Anyway, to prove the lemma in the special case where $x_i=y_i=\left(\sum_{j\ne i}|a_{ij}|^2\right)^{1/2}$, we suppose that $i=1$ (the proof is similar for other values of $i$). Let $A=\pmatrix{a&\mathbf v^T\\ \mathbf v&S}$ and $B=A-aI$.
Since $B$ is symmetric, its singular values are the absolute values of its eigenvalues. As the smallest singular value of $B$ is bounded above by the norm of the first column of $B$, we obtain $|\lambda(B)|=\sigma_\min(B)\le\|\mathbf v\|_2$ for the smallest-sized eigenvalue $\lambda(B)$ of $B$. It follows that $|\lambda(A)-a|^2\le\|\mathbf v\|_2^2$ for the eigenvalue $\lambda(A)$ of $A$ that is the closest to $a$.
