Let the tall matrix ${\bf J} \in {\Bbb R}^{m \times n}$ (with $m>n$) have full column rank. Note that $\bf J$ is the Jacobian of some invertible transformation. Moreover, let the matrix ${\bf K} \in {\Bbb R}^{m \times m}$ be invertible. Is there a way (e.g., using a generalised inverse) to simplify $ \mathbf{J} \left(\mathbf{J}^\top \mathbf{K}\mathbf{J} \right)^{-1} \mathbf{B} = \mathbf{0}$, where $\mathbf{B}$ is a block matrix of the form $\mathbf{B} = \left( \begin{array}{c} \mathbf{I}_n \\ \mathbf{0}_{(m-n)\times n} \end{array}\right)$?
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to simplify the following expression - what exactly do you accept as a simplification? – Dietrich Burde Apr 23 '25 at 08:41
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Where does the matrix $ {\bf J} \left({\bf J}^\top {\bf K} , {\bf J} \right)^{-1} $ come from? – Rodrigo de Azevedo Apr 23 '25 at 08:50
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Is $\bf J$ the Jacobian of a linear transformation? – Rodrigo de Azevedo Apr 23 '25 at 08:51
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I have completed the question... I know how to explicitly calculate $\mathbf{K}^{-1}$ and $\mathbf{J}$ . I would like something that depends on them... – Vuk Apr 23 '25 at 08:55
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Jes, $\mathbf{J}$ is the Jacobian of a locally invertible transformation (not linear in general) – Vuk Apr 23 '25 at 08:56
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1this might be relevant – Ben Grossmann Apr 23 '25 at 18:45
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If $m$ is close to $n$, there is a relatively inexpensive way to use $K^{-1}$ in the computation of $(J^TKJ)^{-1}$, but it's probably not the kind of "simplification" that you're looking for – Ben Grossmann Apr 23 '25 at 18:49
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1There's also an approach that can be taken using Schur complements – Ben Grossmann Apr 23 '25 at 18:56
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@Ben Grossmann Thanks! I'll try to use the Schur complements since this approach may handle the matrix B... Fingers crossed! – Vuk Apr 23 '25 at 21:55