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Let $ \text{GL}^+_n$ be the group of real $n \times n$ matrices with positive determinant, and consider the matrix absolute value function, $| \cdot | : \text{GL}^+_n \to \text{Psym}$ given by $|A|=\sqrt{A^TA}$.

($\sqrt{}$ is the unique symmetric positive-definite matrix square root).

Can the derivative of $|\cdot |$ (in some fixed direction) explode to infinity when $\det A \to 0$?

If this happens, then there should be some "high-dimension" phenomena, since in dimension $1$, we just have the usual absolute value $1$. (In particular, we should probably look for non diagonal examples).

If we denote $|A|=P(A)$, then $P^2=A^TA$; Differentiating this, we get $$P\dot P + P\dot P = \dot A^T A+ A^T \dot A,$$

and this equation uniquely determines $\dot P$ (it's a Sylvester equation).

I also know that when $A$ is positive-definite, then $dP_A $ is bounded independent of $A$*, at least when $n=2$.

*As long as it is positive-definite.

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  • $\begingroup$ How is the derivative of $|\cdot|$ defined? Is it a matrix-by-matrix derivative and, if so, what does 'explode to infinity in a particular direction' mean and what notation convention is being used for the derivative? Or are the elements of $A$ functions of a variable $t$ with all derivatives being taken w.r.t $t$? $\endgroup$ – Angela Richardson May 13 at 14:09
  • $\begingroup$ Well, the context of the question is of smooth maps between manifolds. (Although here the source domain is an open subset of Euclidean space, and the target can also be considered as a subset of $\mathbb{R}^{n^2}$). So, you can think of it as $n^2$ smooth functions defined on an open subset of $\mathbb{R}^{n^2}$. "In a particular direction" means choosing some fixed unit vector $v$, and evaluating $d| \cdot |_A(v)$ when $\det A \to 0$. $\endgroup$ – Asaf Shachar May 13 at 14:13
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There are no explosions. Let $f:A\in GL_n^+\mapsto\sqrt{A^TA}$.

According to

Derivative (or differential) of symmetric square root of a matrix

$Df_A:H\in M_n\mapsto \int_0^{\infty}\exp(-t\sqrt{A^TA})(H^TA+A^TH)\exp(-t\sqrt{A^TA})dt$. We assume that $A_s\rightarrow A_0$ where $A_s\in GL_n^+$ and $\det(A_0)=0$.

$\textbf{Proposition}$ $Df_{A_s}(H)$ is bounded when $H$ is bounded.

$\textbf{Proof}$. (sketch) It suffices to show that $tr(Df_{A_s}(H))$ is bounded. (I remove the $s$ in the calculation)

$tr(Df_{A_s}(H))=\int_0^{\infty}tr(\exp(-t\sqrt{A^TA})(H^TA+A^TH)\exp(-t\sqrt{A^TA}))dt=$

$\int_0^{\infty}tr(\exp(-2t\sqrt{A^TA})(H^TA+A^TH))dt=tr(\int_0^{\infty}\exp(-2t\sqrt{A^TA})dt(H^TA+A^TH))=$

$tr(1/2(A^TA)^{-1/2}(H^TA+A^TH))=tr(A_s({A_s}^TA_s)^{-1/2}H^T)$.

The last expression is bounded because $A_s({A_s}^TA_s)^{-1/2}$ is orthogonal.

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