How two conditions are satisfied by any normalized vector

Question

The vector $\mathbf{u}_{\lambda}$ is assumed to be normalized.
Then I want to understand How these two conditions are satisfied? $$ \begin{aligned} &\sum_{k} u_{k \kappa}^{*} u_{k \lambda}=\delta_{\kappa \lambda} \\ &\sum_{\kappa} u_{k \kappa}^{*} u_{l \kappa}=\delta_{k l} \end{aligned} $$

For more detail on this problem check [See equation 3 & 3' page of this file]1

Answer 1

$M$ is positive definite, therefore it has orthogonal set of eigenvectors. Since we assume them to be normalized, the matrix $ U = [ \mathbf{u}_1, \mathbf{u}_2, ... , \mathbf{u}_n]$ (we treat columns as vectors) is orthonormal. The first formula (U^*U = I) comes from the definition:

• Vectors are normalized -> ones on diagonal
• Orthogonal -> zeros off diagonal

This makes $U^* = U^{-1}$, therefore $UU^* = I$. Take element-wise complex conjugate and you get the second formula.