Suppose H and E are two $n\times m$ matrices with $n>m$, now I have an equation: $$\begin{equation}H=E\rho \end{equation}$$ where $\rho$ is $m\times m$ since $n>m$, we have moore-penrose invereses $H^{+}$ and $E^{+}$ such that $H^{+}H=I$ and $E^{+}E=I$ . Using these two relation I can recast $H=E\rho$ as $E^{+}H=E^{+}E\times\rho$ which gives $E^{+}H=\rho$, alternatively $H^{+}H=E\times \rho$ that gives $I=H^{+}E\rho$, now $H^{+}E$ is a square matrix and can be invertible so that $(H^{+}E)^{-1}=\rho$, but for pseudo invereses, $(H^{+}E)^{-1}\ne E^{+}H$, so how come we have two representations for $\rho$ ?, I am confused which one is the real solution. Matlab example:
and none of them when multiplied by E give me H:
$(H^+E)^{-1}$ is equal to $E^+H$, provided that $H,E$ are "tall" matrices with full column ranks and $H=E\rho$.
However, if you generate $H$ and $E$ individually at random, there may not exist a matrix $\rho$ that makes $H=E\rho$. This is what happened in your numerical example.