0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.