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I ran into the following problem:

Prove that any arbitrary $m \times n$ matrix $A$, there exists an unique $n \times m$ matrix $P$ that satisfies the following equations: $$\begin{cases} APA = A \\ PAP = P \\ (AP)^T = AP \\ (PA)^T = PA \end{cases}$$

What I've done so far:

  • From equations (1) and (3), we get $(AP)^TA = A \Rightarrow (AP)^T = (AP)^{-1}$ so $AP$ is an orthogonal matrix.
  • Similarly, from equations (2) and (4), $PA$ is also an orthogonal matrix.
  • If $A$ has independent columns, $APA = P^TA^TA = A \Rightarrow P = A(A^TA)^{-1}$ also $APA = AA^TP = A \Rightarrow P = (A^TA)^{-1}A$. But I have no idea what to do when $A^TA$ isn't invertible.
  • Using singular value decomposition, $A = U\Sigma V^T = APA$ implies $P = V\Sigma U^T$

I've read this post here, which I had a difficult time trying to understand it. Hoping for some direction to think; I really wonder how to prove that there is a unique $P$ for any $A$. Any tips help, and thanks in advance.

Edit: now I've got some sleep, and found out that I'm wrong about the first two points.

Uduru
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  • This question might be helpful. – Varun Vejalla Jul 25 '22 at 05:25
  • @VarunVejalla Thanks, let me take a look. – Uduru Jul 25 '22 at 05:38
  • The statement as it stands is false. Consider $A=\pmatrix{1\ i}$ for instance. If $AP$ is symmetric, $P$ must be a scalar multiple of $A^T$. But then $PA=0$ (because $A^TA=0$) and hence $APA=0\ne A$. To correct the statement that you want to prove, you need to either assume that $A$ is a real matrix or change all occurrences of matrix transpose to conjugate transposes. – user1551 Jul 25 '22 at 10:01
  • @user1551 I think from the question it is fairly obvious this is about real matrices (and really about Euclidean vector spaces). In the case of complex vector matrices the transposition should be combined with complex conjugation, as you probably know very well. – Marc van Leeuwen Jul 25 '22 at 10:49
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    This is really a duplicate to the question OP linked to; asking the same question again with "I don't understand the answer given there" is not really helpful. I've given an answer to the linked question and then closed this one; as usual anyone wanting to answer this question is invited to add an answer to the one it is a duplicate of. – Marc van Leeuwen Jul 25 '22 at 10:52

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