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I need to solve the following equation for the matrix $P \in\mathbb{R}^{r\times d}:$

$$ ((PAP^\top)^{-1} P S P^\top (PAP^\top)^{-1})^2 = I_r, $$ where $S$ is a symmetric $d\times d$ matrix, $A$ is a PSD $d\times d$ matrix, and $I_r$ is the identity matrix of dimension $r$.

Is there any easy way to solve this equation?

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    $\begingroup$ What's the context for this question? What have you tried? $\endgroup$
    – David Roberts ♦
    Commented Oct 23, 2021 at 8:52
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    $\begingroup$ The context is the resolution of a system of matrix equations that leads to this equations. I have tried solving the scalar version of this equation. It leads to $p = sa^{-1}$. In the matrix case I don't know how to proceed. $\endgroup$
    – Apprentice
    Commented Oct 23, 2021 at 9:30
  • $\begingroup$ I mean something more detailed, like what research problem does this come from? Trying to solve the scalar version is a small start, but not like what I was expecting. Like eg having a solid go at the 2x3 or 3x2 versions $\endgroup$
    – David Roberts ♦
    Commented Oct 23, 2021 at 20:30
  • $\begingroup$ Some random ideas: 1) you have a matrix squaring to the identity, all its eigenvalues are $\pm1$, and its determinant is $\pm1$ 2) you must have the inequality bound on $d\leq r$, else the equation has no solution (the various $PXP^\perp$ won't have full rank) 3) Is $P$ meant to diagonalise $S$, since $S$ is symmetric hence orthogonally diagonalisable? 4) you can get an equation $\det(PAP^\perp)^2 = \det(PSP^\perp)^4$. 5) I'd be inclined to break this into two equations, namely $X=(PAP^\perp)^{-1}$ and $(XPSP^\perp X)^2=I$. $\endgroup$
    – David Roberts ♦
    Commented Oct 24, 2021 at 7:07
  • $\begingroup$ 6) In fact, you might as well start with the Ansatz that $(PAP^\perp)^{-1} PSP^{-1} (PAP^\perp)^{-1} = diag(1,\ldots,1,-1,\ldots,-1)$, since you can get all possible solutions from this one by "undiagonalising". 7) Then you can make various assumptions about block structure of $P$, which amounts to choosing clever bases, and then the general case arises by inserting change of basis matrices. 8) you can probably assume that $A$ is actually positive definite, else a zero eigenvalue would break the equation. And so on. $\endgroup$
    – David Roberts ♦
    Commented Oct 24, 2021 at 7:28

1 Answer 1

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$\DeclareMathOperator\diag{diag}\DeclareMathOperator\rank{rank}$Here is how we can construct solutions in $P$.

Necessarily, $r\leq d$, $\rank(P)=r$, $\rank(S)=r_1\geq r$.

There is $Q$ invertible s.t. $QAQ^T=I_d$, $QSQ^T=\diag((\lambda_i)_{i\leq p_1},(-\mu_j)_{j\leq q_1},0_{s})$, where $\lambda_i,\mu_j>0$ and $p_1+q_1=r_1$, $p_1+q_1+s=d$.

Let $p$, $q$ be s.t. $p\leq p_1$, $q\leq q_1$, $p+q=r$. Even if it means changing the ordering of the diagonal of $QSQ^T$, we may assume that the first $p$ elements of this diagonal are $>0$ and the following $q$ are $<0$.

Let $D\in M_{r,d}$ be the "diagonal" matrix $\diag((d_i)_{i\leq r})$.

Then $U=DQAQ^TD^T=\diag((d_i)^2)$, $V=DQSQ^TD^T=\diag((\lambda_i d_i^2),(-\mu_j d_j^2))$.

Finally $U^{-1}VU^{-1}=\diag\left(\left(\dfrac{\lambda_i}{d_i^2}\right)_{i\leq p},\left(\dfrac{-\mu_j}{d_j^2}\right)_{j\leq q}\right)$ and we choose $d_i=\sqrt{\lambda_i},d_j=\sqrt{\vphantom\lambda\mu_j}$.

For $P=DQ$, the considered expression is $\diag(I_p,-I_q)$ and we are done.

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  • $\begingroup$ I think I don't understand the notation. You say to put $d_i = \sqrt{\lambda_i}$, $d_j = \sqrt{\vphantom\lambda\mu_j}$, but aren't $d_i$ and $d_j$ drawn from the same tuple, just with different names for the subscripts? $\endgroup$
    – LSpice
    Commented Nov 13, 2021 at 23:37
  • $\begingroup$ @LSpice , my notation is not very good. The indices $i$ number the first $p$ elements of the diagonal and the indices $j$ number the last $q$. $\endgroup$
    – loup blanc
    Commented Nov 14, 2021 at 10:37

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