1
$\begingroup$

For a set of column vectors $x_1,\dots,x_n$, the identity shows that $\sum_{i=1}^n x_i x_i^T = X^TX$. I can show this by seeing the $(p,q)$ entry of the resulting matrix is $\sum_{i=1}^n (X^T)_{pi}X_{iq} = \sum_{i=1}^n x_{ip} x_{iq}$. Is there a quicker way of seeing this? and, does $xx^T$ have a special name?

1 Answers 1