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Let $A$ be an $m \times n$ matrix such that $\mathrm{rank}(A) < n$. If one were to set the equation

\begin{equation*} A Z = B A, \end{equation*}

what is the condition that matrix $Z$ needs to fulfill such that the above relation is even possible? $Z$ is an $n \times n$ matrix.

Thank you!

slm992
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1 Answers1

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Presumably you want to solve this for the $m \times m$ matrix $B$. On the null space of $A$, the right side is $0$, so $Z$ must map the null space of $A$ into the null space of $A$. If $Z$ satisfies this condition, then you can take $B = A Z A^+$ where $A^+$ is the Moore-Penrose pseudoinverse of $A$. This works because $A^+A$ is the orthogonal projection on the orthogonal complement of the null space of $A$, so for $x$ in that orthogonal complement, $BAx = A Z A^+ A x = A Z x$.

Robert Israel
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  • Thank you for your answer. However, I would like something else. Lets say I know matrix $A$. Then I would like to find out how should matrix $Z$ look like such that the above relationship is possible. I dont necessarily have to obtain the solution for $B$. Could it be that what I want is impossible? – slm992 Jun 03 '20 at 19:16
  • That's what I provided. If $Z$ maps the null space of $A$ into itself, the equation is possible (and I proved that by providing a $B$ that works). If $Z$ does not map the null space of $A$ into itself, the equation is impossible. – Robert Israel Jun 03 '20 at 19:19
  • Thank you very much! :) – slm992 Jun 03 '20 at 19:20