I have $X = (U'WU)^{-1}U'$ to be solved. Suppose $U'$ is $3 \times 7, W$ is $7 \times 7$ positive definite matrix, $U'$ is of rank 3.
So, I transformed $(U'WU)^{-1}U'$ as
$(U'WU)^{-1}U'WU = I\\ XWU = I\\ U'WX' = I\\ (I \otimes U'W) vec(X') = vec(I).\\ $
When I solved $ X = (U'WU)^{-1}U'$ and as the above linear system using R, the answers are slightly different. Can anyone just point out whether there is something wrong with the above logic.
Thanks
1) A standard and the simplest way (if there's nothing better to do) to compute $X$ is first to compute the Cholesky factorization of the SPD matrix $U'WU=LL'$ and then solve the system with multiple right-hand sides $LL'X=U'$. Note that $U'WU$ is $3\times 3$ so this approach is very cheap.
2) With the approach in question, the problem is already in finding $X$ from $XWU=I$, which is not uniquely solvable [any $3\times 7$ matrix $Y$ whose rows are orthogonal to the columns of $WU$ will also satisfy $(X+Y)WU=I$].