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I'd appreciate help finding the inverse of the upper-left triangular (partitioned) matrix

$$ \left[ \begin{array}{ll} \mathbf{K} & \mathbf{P} \\ \mathbf{P}^T & \mathbf{0} \end{array} \right] $$

This matrix occurs frequently in scattered data interpolation with radial basis functions. If it matters, $\mathbf{K}$ is a square matrix, $\mathbf{P}$ is generally not square and $\mathbf{0}$ is the zero matrix.

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Does the blockwise inversion formula help? –  Rahul Aug 5 '11 at 15:20
    
'Just might -- as long as it does not involve taking the inverse of the zero matrix or of $\mathbf{P}$ and $\mathbf{P}^T$. Edit: looks like it does the trick -- thanks! –  Olumide Aug 5 '11 at 15:40
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up vote 2 down vote accepted

To get this off the "unanswered" list, there is the formula (which I also used here):

$$\mathbf{A}^{-1}=\begin{pmatrix}\mathbf{E}^{-1}+\left(\mathbf{E}^{-1}\mathbf{F}\right)(\mathbf{H}-\mathbf{G}\mathbf{E}^{-1}\mathbf{F})^{-1}\left(\mathbf{G}\mathbf{E}^{-1}\right)&-\left(\mathbf{E}^{-1}\mathbf{F}\right)(\mathbf{H}-\mathbf{G}\mathbf{E}^{-1}\mathbf{F})^{-1}\\-(\mathbf{H}-\mathbf{G}\mathbf{E}^{-1}\mathbf{F})^{-1}\left(\mathbf{G}\mathbf{E}^{-1}\right)&(\mathbf{H}-\mathbf{G}\mathbf{E}^{-1}\mathbf{F})^{-1}\end{pmatrix}$$

for the block matrix $\mathbf{A}=\begin{pmatrix}\mathbf{E}&\mathbf{F}\\ \mathbf{G}&\mathbf{H}\end{pmatrix}$.

Applying this formula to your matrix yields

$$\begin{pmatrix}\mathbf{K}^{-1}-\left(\mathbf{K}^{-1}\mathbf{P}\right)(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\left(\mathbf{P}^\top\mathbf{K}^{-1}\right)&\left(\mathbf{K}^{-1}\mathbf{P}\right)(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\\(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\left(\mathbf{P}^\top\mathbf{K}^{-1}\right)&-(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\end{pmatrix}$$

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