Penrose-Moore Inverse is a special case of the Right-Inverse of a matrix?

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In this book (Alazard) it is said that a right inverse matrix $T^{\#}$ of a generic matrix $T$ may be written as:

$ T^{\#} = T^{+} + T^{\perp}X $

Where $T^{\perp}$ is an orthonormal basis of the null space of $T$ and $X$ is a matrix of free parameters.

I never came across such a general definition of a right inverse (I'm an engineer, so please pardon me!). Where can I find an explanation of this? In my narrow view the right inverse was simply the Penroose-Moore pseudo-inverse and this general form is totally new.

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$A$ being a right inverse of $T$ just means that $TA=I$ where $I$ is the identity matrix on the codomain of $T$. A given matrix may or may not have a right inverse.

The Moore-Penrose inverse $T^+$ of a right-invertible matrix is indeed a right inverse, but the Moore-Pensrose inverse is defined for all matrices, including those that have no right-inverse, thus it is not a special case of the right-inverse.

To see that for a right-invertible matrix, $T^\#$ is indeed a right inverse, you can just calculate: $$TT^\# = T(T^+ + T^\bot X) = TT^+ + TT^\bot X = I +0X = I$$ Note that this uses the relation $TT^+=I$ which obviously is only true for right-invertible matrices.

Conversely if $T^+$ and $S$ are both right inverses of $T$, then $T(S−T^+)=0$, hence $S−T^+$ takes its values from the kernel, and thus it can be written as $T^\bot X$ for some matrix $X$.