Step in proof of Schur Product Theorem

Question

Proof taken from http://www.math.ucsd.edu/~njw/Teaching/Math271C/Lecture_03.pdf, top of page 3.

$A \circ B = \sum_{i,j}w_{ij}w_{ij}^T$, where $w_{ij} = u_i \circ u_j$. $\circ$ is Hadamard product.

I'm confused as to why $w_{ij} = u_i \circ u_j$.

Answer 1

This follows from a direct calculation using $(u u^T)_{ij} = u_i \cdot u_j$ where $u$ is a $n \times 1$ vector:

$$ (w_{ij} w_{ij}^T)_{kl} = (w_{ij})_k \cdot (w_{ij})_l = (u_i \circ u_j)_k \cdot (u_i \circ u_j)_l = (u_i)_k \cdot (u_j)_k \cdot (u_i)_l \cdot (u_j)_l, \\ ((u_i u_i^T) \circ (u_j u_j^T))_{kl} = (u_i u_i^T)_{kl} (u_j u_j^T)_{kl} = (u_i)_k \cdot (u_i)_l \cdot (u_j)_k \cdot (u_j)_l.$$