Inverse of an upper-left triangular (partitioned) matrix 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.
 A: The answer above assumes the block $(1,1)$ of $A$ is nonsingular.
In the case of interpolation by radial basis functions, you do not have this necessarily.
So, you need a more general formula, given by Gansterer.
Look for it in a PDF text by Benzi and Golub on numerical solutions to saddle-point problems, easy to find on Google.
A: 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
$$\begin{pmatrix}\mathbf{K}^{-1}-\left(\mathbf{K}^{-1}\mathbf{P}\right)(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\left(\mathbf{P}^\top\mathbf{K}^{-1}\right)&\left(\mathbf{K}^{-1}\mathbf{P}\right)(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\\(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\left(\mathbf{P}^\top\mathbf{K}^{-1}\right)&-(\mathbf{P}^\top\mathbf{K}^{-1}\mathbf{P})^{-1}\end{pmatrix}$$
