# Pseudoinverses giving weird results

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: • 'i' before 'e', except after 'c' and in weird words like "weird" :-) Feb 9, 2020 at 9:35
• The standard notation for the Moore-Penrose pseudoinverse of a matrix $H$ is $H^+$, and people don't use a $\times$ symbol for matrix multiplication. Feb 9, 2020 at 9:37

$$(H^+E)^{-1}$$ 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.
• @Kutsit Yes.$\phantom{}$. And "full" is an adjective for "rank", not "column". Feb 9, 2020 at 9:58
• @Kutsit That they each has linearly independent columns doesn't mean that they have the same column space. E.g. both $H=\pmatrix{1\\ 0}$ and $E=\pmatrix{0\\ 1}$ have full column ranks, but you cannot find a $1\times1$ matrix (i.e. a scalar) $\rho$ such that $H=E\rho$. Feb 9, 2020 at 10:04
• @Kutsit In fact, if you generate $H$ and $E$ randomly and $n>m$, the probability that there exist a matrix $\rho$ that makes $H=E\rho$ is practically zero. Feb 9, 2020 at 10:08