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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:

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and none of them when multiplied by E give me H:

enter image description here

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    $\begingroup$ 'i' before 'e', except after 'c' and in weird words like "weird" :-) $\endgroup$
    – joriki
    Feb 9, 2020 at 9:35
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    $\begingroup$ 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. $\endgroup$
    – user1551
    Feb 9, 2020 at 9:37

1 Answer 1

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$(H^+E)^{-1}$ 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.

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  • $\begingroup$ Do you mean linearly independent columns by full-column rank ? $\endgroup$
    – Kutsit
    Feb 9, 2020 at 9:56
  • $\begingroup$ @Kutsit Yes.$\phantom{}$. And "full" is an adjective for "rank", not "column". $\endgroup$
    – user1551
    Feb 9, 2020 at 9:58
  • $\begingroup$ But the columns for E and H are LI in the random matrices generated here $\endgroup$
    – Kutsit
    Feb 9, 2020 at 10:01
  • $\begingroup$ @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 $H=E\rho$. $\endgroup$
    – user1551
    Feb 9, 2020 at 10:04
  • $\begingroup$ @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. $\endgroup$
    – user1551
    Feb 9, 2020 at 10:08

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