Suppose, I have a matrix $P\in\mathbb{R}^{m\times n}$ with $\text{rank}(P)=m$ and I'm searching for a right pseudoinverse $P^{+R}$.

Since I'm working with symbolic matrices in a computer algebra system, simply taking a Moore-Penrose-inverse can lead to more complicated inverses. Instead I came up with the following idea:

  1. I can always rearrange the columns of $P$ such that $\tilde{P}=PR=\left(A \big\vert B\right)$ with $A\in\mathbb{R}^{m\times m}$ and $\text{rank}(A)=m$, while $R$ is a permutation matrix.

  2. The right pseudo inverse can then be calculated with $$ P^{+R}= R \begin{pmatrix} A^{-1}\\ 0 \end{pmatrix}. $$

This seems to solve my problem excellently. However, since this is no magic i was wondering if you know any references on this approach?


1 Answer 1


I think you mean $P^{+R} = R \pmatrix{A^{-1}\cr 0\cr}$, so $P P^{+R} = (A |B) R ^{-1} R \pmatrix{A^{-1}\cr 0\cr} = I$.

More generally, this will work for any $n \times n$ invertible matrix $R$ such that the first $m$ columns of $P R$ have rank $m$.

  • $\begingroup$ sure, thanks. i was in a little hurry and just corrected that. still, i guess there must be some literature about that specific approach.. $\endgroup$
    – emma
    Commented Apr 7, 2015 at 19:57

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