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I’m trying to understand Theorem $2.1$ from Fillmore & Williams$^\color{magenta}{\star}$ and I’m a bit confused about the notation used. In particular, the theorem involves an expression of the form $A = B C$. Is the inverse $B^{-1}$ here meant to be the usual matrix inverse, implying that $B$ is square and invertible? Or, is this instead referring to a generalized inverse, such as the Moore-Penrose pseudoinverse, especially if $B$ is not square or not full-rank?

It seems to me that the paper doesn't explicitly clarify this point. Could someone familiar with this type of setup or similar literature shed some light on this? Any intuition or references would also be appreciated.

$\color{magenta}{\star}$ P.A. Fillmore, J.P. Williams, On operator ranges, Advances in Mathematics, Volume 7, Issue 3, December 1971.

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  • $\begingroup$ In which part of Theorem 2.1 does $B^{-1}$ appear? I don't see it. $\endgroup$ Commented Apr 20 at 4:40
  • $\begingroup$ The existence of $C$ is hinged on $B^{-1}$, right? $\endgroup$ Commented Apr 20 at 4:41
  • $\begingroup$ $C = {B_0\!\!}^{-1}A$, where $B_0$ is the restriction of $B$ to $N(B)^\perp$. That's what you mean? $\endgroup$ Commented Apr 20 at 4:43
  • $\begingroup$ Yeah, I do not quite follow that. Pardon me for my "non-math" background, could you explain what it mean? $\endgroup$ Commented Apr 20 at 4:44

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In Theorem 2.1, $A$ and $B$ are bounded operators on a Hilbert space $\mathscr H$; they are not matrices.

In the proof, they define $B_0 \colon \mathscr N(B)^\perp \to \mathscr H$ to be the restriction of $B \colon \mathscr H \to \mathscr H$ to $\mathscr N(B)^\perp$, the orthogonal complement of $\mathscr N(B)$. Note that \begin{align} \mathscr N(B_0) &= \{x \in \mathscr N(B)^\perp : Bx=0\} \\ &= \mathscr N(B)^\perp \cap \mathscr N(B) \\ &= \{0\} \end{align} and that \begin{align} \mathscr R(B_0) &= \{Bx : x \in \mathscr N(B)^\perp\} \\ &= \{Bx : x \in \mathscr H\} & [\text{since } \mathscr H = \mathscr N(B) + \mathscr N(B)^\perp] \\ &= \mathscr R(B). \end{align} So, $B_0$ is injective and has range $\mathscr R(B)$.

Therefore, $B_0 \colon \mathscr N(B)^\perp \to \mathscr R(B)$ is bijective, and then ${B_0\!\!}^{-1} \colon \mathscr R(B) \to \mathscr N(B)^\perp$ (the usual inverse) exists. Note: There is no $B^{-1}$ in the proof.

Now, take $x \in \mathscr H$ and note that $Ax \in \mathscr R(A) \subset \mathscr R(B)$; so ${B_0\!\!}^{-1}Ax \in \mathscr N(B)^\perp$, and then $$ B{B_0\!\!}^{-1}Ax = B_0{B_0\!\!}^{-1}Ax = Ax. $$ Thus, if $C = {B_0\!\!}^{-1}A$, then $$ BCx = Ax \quad \text{for all } x \in \mathscr H. $$ Hence, $BC=A$.

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  • $\begingroup$ Thanks @azif00. With $A=BC$ and $C = B_{0}^{-1}A$ we have $A=B B_{0}^{-1} A$. To a laymann like me, this give $B B_{0}^{-1} = I$ and hence $B = B_{0}$. I am sure I'm missing a lot here. Kindly simplify this for me. $\endgroup$ Commented Apr 20 at 6:32
  • $\begingroup$ @Mike I updated the answer. Is there an specific term you don't understand? $\endgroup$ Commented Apr 20 at 6:48
  • $\begingroup$ Thanks @azif00. Could you shed some light on "what is the motivation on defining $B_{0}: \mathcal{N}^{\perp}(B) \to \mathcal{H}$? I mean why is $ \mathcal{N}^{\perp}(B)$ coming into the picture here? $\endgroup$ Commented Apr 20 at 9:32

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