Certain neighbourhood of diagonal elements of a Hermitian matrix contains at least one eigenvalue of the matrix. Suppose $A=[a_{ij}]$ is an $n \times n$ Hermitian matrix. For each $i, 1 \leq i \leq n$, define 
\begin{align*}
r_{i}= \sqrt{\Sigma_{j \neq i} |a_{ij}|^{2} }
\end{align*}
Then, $[a_{ii}-r_{i}, a_{ii}+r_{i}]$ contains at least one eigenvalue of $A$.  I tried using Spectral theorem and tried to get a contradiction by assuming otherwise but could not get any. I have been trying since a long time. Any hint is appreciable.
 A: As I commented, this link contains an answer. Also Terence Tao's 254A note will be helpful.
A few facts that can help understanding the answer include:


*

*Hermitian matrices have real eigenvalues.

*IF $A$ is a Hermitian matrix, then there is a unitary matrix $U$ and a diagonal matrix $D$ such that $A=UDU^*$. 

*By 1, we can arrange the eigenvalues of a Hermitian matrix $A$ in a non-increasing order, say:
$$\lambda_1(A)\geq \cdots \geq \lambda_n(A).$$

*If $A$ is Hermitian, then the operator norm $||A||$ satisfies
$$
||A||=\max (|\lambda_1(A)|, |\lambda_n(A)|).
$$
All of the above follow from the Spectral Theorem. 


*If $A$ is Hermitian, then for any $1\leq i\leq n$, 
$$
\lambda_i(A) = \sup_{\mathrm{dim}V=i} \inf_{v\in V, \ ||v||_2=1} v^* Av.
$$

*If $A$ and $B$ are Hermitian, then we have
$$
\lambda_i(A) - ||B|| \leq \lambda_i(A+B) \leq \lambda_i(A)+||B||.
$$
We can prove 6. by using 4. and 5. 
Let $A$ be $n\times n$ Hermitian matrix. Now, the answer in the link starts with defining $c_j$ for each $1\leq j\leq n$, where
$c_j$ is the $j$-th column of $A$ in which $j$-th component set to $0$. Then define
$$
E_j = c_j e_j^T + e_j c_j^*$$
where $e_j$ is a standard basis vector. 
Then we obtain the following three facts:
A1. $E_j$ is Hermitian.
A2. $A-E_j$ is Hermitian and it has $a_{jj}$ as an eigenvalue. 
A3. $||E_j||=||c_j||_2= r_j $. 
Find $1\leq i \leq n$ such that $a_{jj}=\lambda_i(A-E_j)$. Then by 6. and A3, we have
$$
|\lambda_i(A)-a_{jj}| \leq ||E_j|| = ||c_j||_2= {r_j},
$$
equivalently $\lambda_i(A)\in [a_{jj}- {r_j}, a_{jj}+ {r_j}]$.
This completes the proof.
