There is a theorem says that:
Let $A = [a_{ij}] \in M_n$ be Hermitian. If $a_{ii} = \lambda_{1}$, then $a_{ik} = a_{ki} = 0$ for all $k=1,2,\ldots, n , k\neq i.$ The eigenvalues are ordered as $\lambda_{1} \leq \lambda_{2} \leq \ldots \leq \lambda_{n}$.
I would like to prove, and this is trivial for the $n= 2$. However I need some hint to prove it.
This question is actually a special case of Rayleigh's principle, which says that for any self-adjoint matrix $A$, its minimum eigenvalue is given by $\min_{\|x\|=1}x^\ast Ax$ and if $v=\arg\min_{\|x\|=1}x^\ast Ax$, then $v$ is an eigenvector corresponding to the minimum eigenvalue of $A$.
Suppose $\lambda_1=\cdots=\lambda_k<\lambda_{k+1}\le\cdots\le\lambda_n$. Let $A=U^\ast\Lambda U$ be a unitary diagonalisation, where $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Let also $u=(u_1,\ldots,u_n)^\top$ be the $i$-th column of $U$. Then one can show that $u^\ast\Lambda u=a_{ii}=\lambda_1$. Hence $u_{k+1}=\cdots=u_n=0$, $u$ is an eigenvector of $\Lambda$. In turn, $e_i$, the $i$-th vector in the standard basis of $\mathbb R^n$, is an eigenvector of $A$.