An inequality relating to entries of eigenvectors

Question

Let $A=[a_{ij}]$ be a $3\times 3$ matrix, where $a_{ii}$ is a real number, and $a_{ji}=\overline{a_{ij}}$ is the complex conjugate of $a_{ij}$ for all $1\leq i,j\leq 3$, i.e $A^t=[\overline{a_{ij}}]$. Let $\lambda_1,\lambda_2,\lambda_3$ be eigenvalues of $A$ (not necessarily distinct) and $u_i=(a_i,b_i,c_i)$ be an eigenvector of $A$ corresponding to $\lambda_i$, $i=1,2,3$. Is this true that $\sum_{i=1}^3\frac{|x_i\overline{y_i}|}{||u_i||^2}\leq 1$, where $x_i,y_i\in\{a_i,b_i,c_i\}$ (not necessarily distinct)? Is this true in general for an $n\times n$ matrix by replacing $n$ with $3$?

Answer 1

The inequality holds for any $n$ if all eigenvalues of the Hermitian matrix $A$ are distinct, so that the eigenvectors form a unitary matrix.$^\ast$

It does not hold if some eigenvalues are identical.
In that case the corresponding eigenvectors need not be orthogonal, you could choose them nearly parallel, say $u_1=(1,0,0)$ and $u_2=(\sqrt{1-\epsilon},\sqrt\epsilon,0)$, with $u_3=(0,0,1)$; then take $x_i=a_i$, $y_i=a_i$, and you find that the inequality is violated: $$\sum_{i=1}^3\frac{|x_i\overline{y_i}|}{||u_i||^2}=2-\epsilon>1.$$

---

$^\ast$ If the eigenvectors $u_i$ are the rows of an $n\times n$ unitary matrix $U$, the inequality in the OP is the statement that $$\sum_{i=1}^n |U_{ij}\bar{U}_{ik}|\leq 1,$$ for any pair of integers $j,k\in\{1,2,\ldots n\}$. Define $U'_{ij}=U_{ij}\,\exp\bigl(-i\arg[U_{ij}\bar{U}_{ik}]\bigr)$, then $$\left(\sum_{i=1}^n |U_{ij}\bar{U}_{ik}|\right)^2=\left(\sum_{i=1}^n U'_{ij}\bar{U}_{ik}\right)^2\leq\left(\sum_{i=1}^n|U'_{ij}|^2\right)\left(\sum_{i=1}^n|U_{ij}|^2\right)=1.$$