Conditional Inverse: Find set of solutions for $AXA=A, A:=\left(1_N-\alpha\iota_N '\right)$.

Question

Given $\alpha \in\mathbb{R}^{N}$ with $\alpha'\iota_N=1$, how can I characterize the set of conditional inverses (or c-inverse) of $\left(1_N-\alpha\iota_N '\right)$ defined as: $\{K\in\mathbb{R}^{N \times N}: \left(1_N-\alpha\iota_N '\right)K\left(1_N-\alpha\iota_N '\right)=\left(1_N-\alpha\iota_N '\right)\}$ where $1_N$ is the identidy matrix of size $N$ and $\iota_N$ is a vector of $1$ of size $N$?

Answer 1

Any matrix $X$ such that $AXA=A$ can be written as $$ X=X_0+Y+Z, $$ where $X_0$ is any particular solution to $AXA=A$ and $Y$ and $Z$ are such that $AY=0$ and $ZA=0$, that is, the columns of $Y$ and the rows of $Z$ lie in the nullspace of $A$ and $A^T$, respectively. If $A=I-bc^T$ such that $c^Tb=1$, we can choose $X_0=A=I-bc^T$ (indeed, $AX_0A=A$). Now looking at the nullspace of $A$, we have $0=Ax=x-(c^Tx)b$ if and only if $x=\xi b$ for some scalar $\xi$, which gives $\mathrm{Ker}(A)=\mathrm{span}(b)$. Similarly, $\mathrm{Ker}(A^T)=\mathrm{span}(c)$. So, $S:=\{X\in\mathbb{R}^{N\times N}:\;AXA=A\}$ is $$ S=\{A+by^T+zc^T:\;y,z\in\mathbb{R}^N\}. $$