 # If G is a generalized inverse of a matrix A (i.e. AGA=A), then is it true that every generalized inverse of A can be written in the form G+B−GABAG for some matrix B of same order as G? Jamar Hays 2022-09-13 Answered
If G is a generalized inverse of a matrix A (i.e. AGA=A), then is it true that every generalized inverse of A can be written in the form G+B−GABAG for some matrix B of same order as G?
I will show that this matrix is a generalized inverse for every matrix B, since
$\begin{array}{rl}A\left(G+B-GABAG\right)A& =AGA+ABA-AGABAGA\\ & =A+ABA-ABA\\ & =A\end{array}$
But I cant conclude that every generalized inverse of A can be written in this form. Help with that
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Let $f\left(X\right)=AXA,\phantom{\rule{thinmathspace}{0ex}}g\left(X\right)=GAXAG$ and $\pi =\mathrm{id}-g$. One can verify that g and in turn $\pi$ are idempotent. Also, $f\pi =0$. Therefore
$\mathrm{range}\left(\pi \right)\subseteq \mathrm{ker}\left(f\right)\subseteq \mathrm{ker}\left(g\right).$
However, as $\pi$ is g are complementary projections to each other, we have $\mathrm{range}\left(\pi \right)=\mathrm{ker}\left(g\right)$. Thus $\mathrm{range}\left(\pi \right)=\mathrm{ker}\left(f\right)=\mathrm{ker}\left(g\right)$ by the sandwich principle.
Now, for any generalised inverse X of A, we have $X-G\in \mathrm{ker}\left(f\right)=\mathrm{range}\left(\pi \right)$. Hence $X=G+\pi \left(B\right)$ for some matrix B.