I am reading Matrix Algebra - Theory, Computations, and Applications in Statistics by James E. Gentle (2007), and I am stuck on an exercise to show that the expression (3.165, p.101) below for a generalized inverse of a partitioned matrix is correct:
This seems similar in form to the inverse of the partitioned matrix, but in this case, there are no restrictions mentioned on A or the partitioned matrices, so A could be non-square.
EDIT: A generalized inverse $A^-$ here refers to any matrix such that $AA^-A = A$
EDIT: Added full name of text and author
I tried but I am unable to extend the derivation of the square partitioned inverse case to this generalized case. I have also tried applying the given expression (3.165) to a simple matrix:
\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}
and obtained a matrix that does not seem to be the generalized inverse (In this case I take $A_{11}$ as the 2$\times$1 matrix):
\begin{bmatrix} -0.5 & 1.166667 & 0.5 \\ -1 & 0.25 & -0.416667 \end{bmatrix}
Is there something I'm missing here, and is there a simple solution to show that the expression is the generalized inverse?
I have no access to the source, and searching for Gentle Matrix Algebra delivers many Gentle Introduction to Matrix Algebra entries...
So i will use the setting from Moore-Penrose inverse which seems to be similar, but the notation differs.
Let us do some computations, but of course in a setting without indices. Let $$ \begin{aligned} M&= \begin{bmatrix} A & B\\ C & D \end{bmatrix}\text{ be a partitioned matrix, associate then} \\ Z &:= D-CA^-B\ ,\text{ and inductively on block shape}\\ M^-&:= \left[ \begin{array}{c|c} A^- + A^-BZ^-CA^- & -A^-BZ^- \\\hline -Z^-CA^- & Z^- \end{array} \right] \ .\text{ Then we have:} \\[3mm] MM^- &= \begin{bmatrix} A & B\\ C & D \end{bmatrix} \left[ \begin{array}{c|c} A^- + A^-BZ^-CA^- & -A^-BZ^- \\\hline -Z^-CA^- & Z^- \end{array} \right] \\ &= \left[ \begin{array}{c|c} AA^- - (I-AA^-)BZ^-CA^- & (I-AA^-)BZ^- \\\hline CA^- - (D-CA^-B)Z^-CA^- & (D-CA^-B)Z^- \end{array} \right] \\ &= \left[ \begin{array}{c|c} AA^- - (I-AA^-)BZ^-CA^- & (I-AA^-)BZ^- \\\hline CA^- - ZZ^-CA^- & ZZ^- \end{array} \right] \ , \\ N:=MM^-M &= \left[ \begin{array}{c|c} AA^- - (I-AA^-)BZ^-CA^- & (I-AA^-)BZ^- \\\hline CA^- - ZZ^-CA^- & ZZ^- \end{array} \right] \begin{bmatrix} A & B\\ C & D \end{bmatrix} \\ &\text{ has then the entries} \\[3mm] N_{11} &= AA^-A - (I-AA^-)BZ^-CA^-A \\ &\qquad\qquad + (I-AA^-)BZ^-C \\ &=A + (I-AA^-)BZ^-C(I-A^-A)\ ,\\[3mm] N_{12} &=AA^-B - (I-AA^-)BZ^-CA^-B \\ &\qquad\qquad + (I-AA^-)BZ^-D\\ &=AA^-B + (I-AA^-)BZ^-Z\ ,\\[3mm] N_{21} &= CA^-A - ZZ^-CA^-A \\ &\qquad\qquad+ ZZ^-C\\ &= CA^-A + ZZ^-C(I-A^-A)\ ,\\[3mm] N_{22} &= CA^-B - ZZ^-CA^-B \\ &\qquad\qquad+ ZZ^-D\\ &= CA^-B + ZZ^-(D-CA^-B)\\ &= CA^-B + ZZ^-Z\\ &= CA^-B + Z\\ &= D\ . \end{aligned} $$
So only the value $D$ is guaranteed. (Note that the value $I$ corresponds to two the one or each other identity matrix, such that $IB$, resp. $CI$ make sense.) We need more. There are some properties that can be used, but i could not find something to have a better comparison. (Please compare the above entries in $N$ with the four entries $A,B,C,D$ in $M$, building the difference. Some (non-commutative) factors occur, and sufficient conditions can be stated, but these are very restrictive for the given generality of the question.)
I was searching for a simple counterexample, so here is my first quick try.
The relation $MM^-M=M$ can not work in general (with my choices of minus-elements, alias generalized matrices.) For this, take $A=0$ (in some block dimension) and $D=0$ (in an other one), so $Z=0$, my choices are then $A^-=0$, $Z^-=0$ (of transposed shapes). Then we compute $$ MM^-M = \begin{bmatrix} 0 & B\\ C & 0 \end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & B\\ C & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \ , $$ which is not $M$ with a high statistical confidence. This was a special case, i know it is not fair to restrict to such a special choice, which is humanly generated. But in the general case we have a very good statistics for the formula.