# Why are these two statements about vector products equivalent?

Let $w_1, \dots, w_m \in \mathbb{C}^d$.

Condition (1) is:

$\sum_i |\langle v, w_i \rangle |^2 = \eta$ whenever $\|v\| = 1$.

Condition (2) is:

$\sum_i u_i u_i^* = I^d$, where $u_i = w_i / \sqrt{\eta}$

This paper claims that the two conditions are equivalent (top of page 3, just beneath the statement of Corollary 1.3). I can't figure out why that would be. Can you help point me in the right direction?

Note that for $\def\norm#1{\left\|#1\right\|}\def\C{\mathbb C}$$v \in \C^d$ with $\norm v = 1$, that $\def\abs#1{\left|#1\right|}\def\sp#1{\left<#1\right>}$ \begin{align*} \sum_i \abs{\sp{v,w_i}}^2 &= \sum_i \overline{\sp{v,w_i}}\\\sp{w_i,v} &= \sum_i \sp{w_i,v}\sp{v,w_i}\\ &= \sum_i v^*w_i w_i^*v\\ &= v^* \sum_i w_i w_i^* \cdot v\\ &= \eta \cdot v^*\sum_i u_i u_i^* \cdot v\\ \end{align*} So the sum is equal to $\eta$ for all such $v$ iff the quadratic form represented by $A = \sum_i u_i u_i^*$ is equal to $v \mapsto \norm v^2$, that is iff $A$ is the identity.