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My question could be considered more general than the $5 \times 5$ case, but I'm wondering how when using Blockwise inversion you can select an appropriate $\bf A$, $\bf C$, and $\bf B$ such that the matrix sizes confirm for multiplication of $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$

For example,

  • $\bf A$ is $2 \times 2$
  • $\bf B$ is $2 \times 3$
  • $\bf C$ is $3 \times 2$
  • $\bf D$ is $3 \times 3$

Lead to $\textbf{A}^{-1}\textbf{B}$ being a $2 \times 3$ matrix which is improper for multiplication by a $3 \times 3$.

Trying $\bf A$ being $3 \times 3$ seemed to lead to the same issue, am I missing something?

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    No it should be OK. $(3 x 2) (2x2) (2x3)$. The internal dimensions must agree and they will do pairwise no matter which order you do the multiplication (2=2 and 2=2). You end up with $3x3$. ($\bf C$ is according to your list not a 3x3 but it will be a $3x2$ multiplied with $2x3$ which gives a $3x3$) – 2017-01-15
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    Schur's complement associated with a $2 \times 2$ partition with square diagonal blocks $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ (it is its name) is always well defined. – 2017-01-15
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    We will need to make sure that any block which is non-square does not have any inverse occuring in the formula as it will be impossible to construct a lossless inverse for any non-square matrix. For example if you had chosen $\bf A$ to be $2x3$ you would have trouble finding ${\bf A}^{-1}$. Except for that all the size-considerations should be compatible no matter which partition you make. – 2017-01-15

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