# 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.

## 1 Answer

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 $-A$.