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.

  • $\begingroup$ 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 $t$? $\endgroup$ May 13, 2019 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 $\det A \to 0$. $\endgroup$ May 13, 2019 at 14:13

1 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)




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


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

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

  • $\begingroup$ Thank you very much for this answer. I apologize for my late response; I somehow missed it until now... I have two questions: (1) Why it suffices to show that $tr(Df_{A_s}(H))$ is bounded? We only know that $tr(Df_{A_s}(H))$ is symmetric, so a-priori it can have very large eigenvalues with different signs. (Do you rely on the fact that you can choose different "test" $H$'s, thus somehow taking care of the signs? Can you elaborate on that?). (2) I don't see why $tr(\int_0^{\infty}\exp(-2t\sqrt{A^TA})dt(H^TA+A^TH))=tr(1/2(A^TA)^{-1/2}(H^TA+A^TH))$... $\endgroup$ Jul 9, 2019 at 1:40
  • $\begingroup$ I do see why $tr(\int_0^{\infty}\exp(-2t\sqrt{A^TA})dt=tr(1/2(A^TA)^{-1/2})$, but this is not exactly what you are claiming. Can you please elaborate on this asserted equality? $\endgroup$ Jul 9, 2019 at 1:40

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