# Where can I find more information about this kind of "pseudo-inverse"?

I have the following matrix: $$A = \left[ \begin{array}{llll} +1 &-1 &+0 &+0\\ +0 &+1 &-1 &+0\\ +1 &+0 &+0 &-1\\ \end{array}\right]$$ for which I verified that the typical right pseudo-inverse $$A^{\dagger}=A^T(AA^T)^{-1}$$ is $$A^{\dagger} = \left[ \begin{array}{llll} +0.50 &+0.25 &+0.25\\ -0.50 &+0.25 &+0.25\\ -0.50 &-0.75 &+0.25\\ +0.50 &+0.25 &-0.75\\ \end{array}\right]$$ which verifies $$AA^{\dagger}=I_{3\times 3}$$ with $$I_{3\times 3}$$ the $$3\times 3$$ identity matrix.

However, I (manually) found that the matrix $$M= \left[ \begin{array}{llll} 2&2&2 \\ 1&2&2 \\ 1&1&2 \\ 2&2&1 \end{array}\right]$$

also, satisfies $$AM=I_{3\times 3}$$, and I cant find any relation between $$A^{\dagger}$$ and $$M$$. Does anyone know what exactly is $$M$$ with respect to $$A$$? Is there another technique to compute a different pseudo-inverse (which will have $$M$$ as its output) that I am not aware of? am I missing something?

Thanks in advance.

• Note that finding a right inverse involves solving a system of $9$ linear equations in $12$ unknowns. Hence, there are degrees of freedom. Apr 4 '20 at 13:48
• You are totally right. There are infinite solutions, and both matrices satisfy that system of equations (As I have verified). It was as simple as that. Thanks. Apr 4 '20 at 14:04
• Of the matrices that satisfy $A X = I$, $A^{\dagger}$ has the smallest Euclidian norm. Apr 4 '20 at 14:09
• You may want to take a look at this. Apr 4 '20 at 14:09
• Thanks guys. Those comments clarified a lot to me. This makes sense now. Apr 4 '20 at 14:19

## 1 Answer

As pointed out by a user comment, $$M$$ is just a right inverse of $$A$$. Provided that $$B$$ is a right inverse of $$A$$, since $$AM=I\Rightarrow A(M-B)=0$$, every right inverse of $$A$$ is in the form $$M=B+K$$, where $$K$$ is any matrix such that $$AK=0$$ (i.e. $$K$$ is any matrix whose columns lie in the null space of $$A$$). In your case, as $$AA^+$$ is indeed equal to $$I$$, all right inverses of $$A$$ are in the form of $$A^++K$$ for some matrix $$K$$ such that $$AK=0$$.

Nicholas Higham happened to have blogged about generalised inverse earlier today. You may learn more about Moore-Penrose pseudoinverse from his blog entry.