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Sorry if there are any formatting issues with my question. This is my first time asking one here. Suppose $A$ is a full column rank matrix that is $m\times n$ (in other words A is long and skinny). Have $A$ be such that

$\mathbf A=\bigl[\begin{smallmatrix}\mathbf B\\\ C\end{smallmatrix}\bigr]$

where $B$ is a $n\times n$ top block matrix that is itself invertible (it is made of $n$ linearly independent rows of $A$), and $C$ is the block that holds the rest of the matrix. Is there any way to find the first $n\times n$ block of the pseudo inverse of $A$. In other words, if

$ A^+ = \begin{bmatrix}\mathbf{U} & \mathbf{V}\end{bmatrix}$

where $U$ is the first $n\times n$ block and $V$ is the rest of the $n\times m$ pseudo inverse of $A$ is there a formula or iterative method to only find this first block $U$ (without computing the entire pseudo inverse). A similar post to this lists some good resources but did not have what I was looking for Blockwise Moore-Penrose pseudoinverse?. In particular, one of the answers suggests that this formula can be found in the Hall and Hartwig paper http://dx.doi.org/10.1137/0130056, but this mostly addresses the 2 by 2 case (or it is going over my head). Also, there seems to be a way to do this partitioning based on columns instead of rows (https://en.wikipedia.org/wiki/Block_matrix_pseudoinverse and https://epubs.siam.org/doi/abs/10.1137/0112050), so I think this is possible. Thanks in advance.

AllCatsAreGrey
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Flash
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1 Answers1

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Maybe this is simpler than what you are looking for. (I'd put this as a comment asking for clarification but I don't have the rep.)

For this case,

$$A^{+}=(A^{\ast }A)^{-1}A^{\ast }$$

(e.g.,The Matrix Cookbook, bottom p 22 https://www2.imm.dtu.dk/pubdb/edoc/imm3274.pdf).

So in partitioned form

$$\begin{bmatrix} U & V\\ \end{bmatrix} = \left( \begin{bmatrix} B^{\ast } & C^{\ast }\\ \end{bmatrix} \begin{bmatrix} B \\ C \\ \end{bmatrix}\right)^ {-1} \begin{bmatrix} B^{\ast } & C^{\ast } \\ \end{bmatrix}$$

$$=\left( B^{\ast }B+C^{\ast }C\right) ^{-1} \begin{bmatrix}B^{\ast } & C^{\ast }\\ \end{bmatrix} $$ $$= \begin{bmatrix} \left( B^{\ast }B+C^{\ast }C\right) ^{-1}B^{\ast } & \left( B^{\ast }B+C^{\ast }C\right) ^{-1}C^{\ast }\\ \end{bmatrix} $$

and so the first block is $U=\left( B^{\ast }B+C^{\ast }C\right)^{-1}B^{\ast }$

dharr
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  • Thank you so much. This is pretty much exactly what I was looking for. If their is another faster way to get this first block that would be cool too, but a method like that might not exist. – Flash Sep 11 '24 at 00:02