Suppose $Z\in \mathbb R^{n\times k}$ and $S=Z^TZ$. Let now $Z(i, x)$ be the matrix $Z$ where the $i-th$ column has been replaced with $x$.
Given $S^{-1}$ is there a quick way to compute $(Z(i,x)^TZ(i,x))^{-1}$?
Suppose $Z\in \mathbb R^{n\times k}$ and $S=Z^TZ$. Let now $Z(i, x)$ be the matrix $Z$ where the $i-th$ column has been replaced with $x$.
Given $S^{-1}$ is there a quick way to compute $(Z(i,x)^TZ(i,x))^{-1}$?
Copyright © 2021 JogjaFile Inc.
The matrix $Y:=Z(i,x)$ can be expressed as $Y=Z+ye_i^T$, where $e_i$ is the $i$th column of the $k\times k$ identity matrix, $y:=x-z_i$, and $z_i$ is the $i$th column of $Z$. If you set $u:=v+1/2\|y\|_2^2e_i$, where $v:=Z^Ty$, then $$ Y^TY=Z^TZ+e_iw^T+we_i^T=Z^TZ+[e_i,w][w,e_i]^T $$ is the rank-two update of $Z^TZ$. Now you can use the Woodbury identity.