How would Moore-Penrose pseudoinverse look like for a simple general example of a block matrix?

$$ A_1=\begin{pmatrix} X & 0 \\ 0 & 0 \\ \end{pmatrix} $$ $$ A_2=\begin{pmatrix} 0 & X \\ 0 & 0 \\ \end{pmatrix} $$

Where X is a block matrix.

I tried some online calculator and got:

$$ A^+_1=\begin{pmatrix} X^+ & 0 \\ 0 & 0 \\ \end{pmatrix} $$ $$ A^+_2=\begin{pmatrix} 0 & 0 \\ X^+ & 0 \\ \end{pmatrix} $$

Where $A_i^+$ and $X^+$ denote the pseudoinverses.

But why there is the flip for $A_2$? Because of the transposition?

Is there any general formula how to deal with the block matrices like this?

On Wikipedia I just found the block matrices written in the form of $A = (X | Y)$ or the general expression for the inverse of the block matrices but I wasnt able to find out why that flip actually happened in these formulae.

And what if $X$ isn't regular or square?


1 Answer 1


Yes, because of the transposition.

Recall that $A^+$ is the unique matrix such that $AA^+A=A,\,A^+AA^+=A^+$ and both $AA^+$ and $A^+A$ are Hermitian. So, if $P$ is a permutation such that the product $AP$ makes sense, then $(AP)(P^TA^+)(AP)=AP,\,(P^TA^+)(AP)(P^TA^+)=P^TA^+$ and both $(AP)(P^TA^+),\,(P^TA^+)(AP)$ are Hermitian. It follows that $(AP)^+=P^TA^+$. That is, if you permute the columns of $A$, the rows of $A^+$ will be permuted too.

There is actually a special case of the more general result that $(AB)^+=B^+A^+$ when $B$ has orthonormal rows.


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