Matrix decomposition into inverse format $(AX)^{-1}A$

Question

Assume $X$ to be $n \times k$ and $H$ to be $k \times n$.

Given that $HX = I$, can I prove that $H$ can be decomposed into the form $(AX)^{-1}A$?

$(AX)^{-1}A$ implies $HX = I$, but would the converse necessarily be true? And how can I prove it?

Answer 1

Assuming $H$ cannot be decomposed then there is no such form $(AX)^{-1}A$ such that $(AX)^{-1}AX = I$

Also, $A$ must be $k \times n$ so that $(AX)^{-1}$ is $k \times k$ invertible.

Therefore, $A$ can be equated to $A=BH\;\;\;$ ($B$ - some matrix $k \times k$)

$(BHX)^{-1}(BH)=(BI)^{-1}(BH)=B^{-1}(BH)=(B^{-1}B)H=IH=H$