Can the derivative of the matrix absolute value explode when we approach singular matrices?

Question

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.

Answer 1

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.

EDIT 1. Answer to Asaf Shachar.

i) We want to show that the $\dfrac{\partial f}{\partial {A_s}_{k,l}}$ are bounded by an expression that does not depend on $s,k,l$. Let $H=[h_{k,l}]$ with $h_{i,j}=1$ and the other are $0$. Then $tr(Df_{A_s}(H))=\dfrac{\partial tr(f)}{\partial {A_s}_{i,j}}$; indeed, it seems that bounding the trace is not enough to solve the problem.

ii) $\int_0^{\infty}\exp(-2t\sqrt{A^TA})dt=[-1/2(A^TA)^{-1/2}\exp(-2t\sqrt{A^TA})]_0^{+\infty}=$

$1/2(A^TA)^{-1/2}$.

EDIT 2. Perhaps, it is easier to show that $f$ is Lipschitz.

We may assume that $A_s^TA_s$ tends to $A_0^TA_0=diag((\sigma_i^2)_i)$ where at least one $\sigma_i$ is $0$. Then $\sqrt{A_s^TA_s}$ tends to $\sqrt{A_0^TA_0}=diag((\sigma_i)_i)$. It "remains" to show that

$||\sqrt{A_s^TA_s}-diag((\sigma_i)_i)||_2\leq ||A_s-A_0||_2.$