 # Transformation of inverse to a system of linear equations. Need to solve X=(U′WU)−1U′. U′ is U′ is 7x7 positive definite matrix, U′ is of rank 3 Pavukol 2022-09-11 Answered
Transformation of inverse to a system of linear equations. Need to solve $X=\left({U}^{\prime }WU{\right)}^{-1}{U}^{\prime }$. ${U}^{\prime }$ is ${U}^{\prime }$ is $7×7$ positive definite matrix, ${U}^{\prime }$ is of rank $3$.
Transformed $\left({U}^{\prime }WU{\right)}^{-1}{U}^{\prime }$ as
$\left({U}^{\prime }WU{\right)}^{-1}{U}^{\prime }WU=I\phantom{\rule{0ex}{0ex}}XWU=I\phantom{\rule{0ex}{0ex}}{U}^{\prime }W{X}^{\prime }=I\phantom{\rule{0ex}{0ex}}\left(I\otimes {U}^{\prime }W\right)vec\left({X}^{\prime }\right)=vec\left(I\right).\phantom{\rule{0ex}{0ex}}$
When I solved $X=\left({U}^{\prime }WU{\right)}^{-1}{U}^{\prime }$ and as the above linear system using $R$, the answers are slightly different. Does something wrong with the above logic?
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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 ${U}^{\prime }WU=L{L}^{\prime }$′ and then solve the system with multiple right-hand sides $L{L}^{\prime }X={U}^{\prime }$. Note that ${U}^{\prime }WU$ is $3×3$ 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 $3×7$ matrix $Y$ whose rows are orthogonal to the columns of $WU$ will also satisfy $\left(X+Y\right)WU=I$.