Eigenvalues on the diagonal of a Hermitian(self-adjoint) matrix. I appreciate any help that can be given on this, I just can't seem to get started.
Let $A$ be an $n\times n$ Hermitian matrix with eigenvalues $\lambda_1 \leq \lambda_2\leq \cdots \leq \lambda_n$.  Prove that if $a_{ii}=\lambda_1$ for some $i$, then every other entry of row and column $i$ is zero.  Similarly, prove that if $a_{ii}=\lambda_n$ for some $i$, then every other entry of rows and column $i$ is zero.
Way back when, we had the theorem "The eigenvalues of a triangular matrix are the entries on its main diagonal."  So this makes since to me,but I doubt very much that is useful in this case.  This is from a chapter on Hermitian matrices and have theorems such as Courant-Fischer and Cauchy's Interlacing which seem to draw some similarities, but honestly I just have no clue.
 A: I'm not claiming this is the easiest proof, but it is what came to mind. 
Write $e_1,\ldots,e_n$ for the canonical basis. 
Since $A$ is selfadjoint, we can write $$ A=\sum_{j=1}^n\lambda_j P_j,$$ where $P_1,\ldots,P_n$ are (pairwise orthogonal)  rank-one projections that add to the identity. 
Now, if $A_{ii}=\lambda_1$, 
$$
\lambda_1=A_{ii}=\langle Ae_i,e_i\rangle=\sum_{j=1}^n\lambda_j\langle P_je_i,e_i\rangle\geq\sum_{j=1}^n\lambda_1\langle P_je_i,e_i\rangle
=\lambda_1\langle\sum P_je_i,e_i\rangle=\lambda_1\langle e_i,e_i\rangle=\lambda_1. 
$$
Then the inequality is an equality, and we have a sum of non-negative terms
$$
\sum_{j}(\lambda_j-\lambda_1)\langle P_je_i,e_i\rangle=0.
$$
If the eigenvalues are $\lambda_1,\ldots,\lambda_1,\lambda_k,\lambda_{k+1},\ldots,\lambda_n$, for $j\geq k$ we have  $\lambda_j>\lambda_1$, and so we get $\langle P_je_i,e_i\rangle=0$. Let $Q=\sum_{j\geq k}P_j$. We have, since $Q $ is a projection, $$0=\langle Qe_i,e_i\rangle=\langle Qe_i,Qe_i\rangle,$$
so $Qe_i=0$. Then $(P_1+\ldots+P_{k-1})e_i=e_i$. In particular, $Ae_i=\lambda_1e_i$.
Now, for $j\ne i$,
$$
A_{ij}=\langle Ae_j,e_i\rangle=\langle e_j,Ae_i\rangle=\lambda_1\langle e_j,e_i\rangle=0.
$$
As $A$ is selfadjoint, $A_{ji}=\overline{A_{ij}}=0$. 
The case for $A_{ii}=\lambda_n$ follows from the previous one by working with $-A$. 
