Transformation of inverse to a system of linear equations. Need to solve X = ( U ′ W U ) − 1 U ′ . U ′ is U ′ is 7 × 7 positive definite matrix, U ′ is of rank 3.Transformed ( U ′ W U ) − 1 U ′ as ( U ′ W U ) − 1 U ′ W U = I X W U = I U ′ W X ′ = I ( I ⊗ U ′ W ) v e c ( X ′ ) = v e c ( I ) . When I solved X = ( U ′ W U ) − 1 U ′ and as the above linear system using R, the answers are slightly different. Does something wrong with the above logic?

Question

Transformation of inverse to a system of linear equations. Need to solve   X  =  (      U    ′    W  U      )          −      1            U    ′  .       U    ′   is       U    ′   is   7  ×  7 positive definite matrix,       U    ′   is of rank   3.Transformed   (      U    ′    W  U      )          −      1            U    ′   as  (      U    ′    W  U      )          −      1            U    ′    W  U  =  I    X  W  U  =  I        U    ′    W      X    ′    =  I    (  I  ⊗      U    ′    W  )  v  e  c  (      X    ′    )  =  v  e  c  (  I  )  .  When I solved   X  =  (      U    ′    W  U      )          −      1            U    ′   and as the above linear system using   R, the answers are slightly different. Does something wrong with the above logic?

faliryr · Accepted Answer

1) A standard and the simplest way (if there&#039;s nothing better to do) to compute   X is first to compute the Cholesky factorization of the SPD matrix       U    ′    W  U  =  L      L    ′  ′ and then solve the system with multiple right-hand sides   L      L    ′    X  =      U    ′  . Note that       U    ′    W  U is   3  ×  3 so this approach is very cheap.2) With the approach in question, the problem is already in finding   X from   X  W  U  =  I, which is not uniquely solvable [any   3  ×  7 matrix   Y whose rows are orthogonal to the columns of   W  U will also satisfy   (  X  +  Y  )  W  U  =  I.

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

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