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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.

FeedbackLooper
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1 Answers1

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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.

user1551
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