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Let $B\in \mathbb{R}^{n\times n}$ be a positive semi-definite matrix (i.e., $B \in \mathbb{S}^n_{\ge 0}$), and consider the map $$ \mathbb{R}^{n\times n}\ni A \mapsto f(A):=(AA^\top +B)^{1/2}\in \mathbb{S}^n_{\ge 0}, $$ where $(\cdot)^{1/2}$ is unique positive semidefinite matrix square root. Is it possible to prove the map is Lipschitz continuous? How about local Lipschitz continuity?

The answer is affirmative for one-dimensional case. By discussing whether $B=0$ or not, one can easily show $f(A)$ is Lipschitz continuous.

In a multidimensional setting, if $B$ is positive definite, then the matrix square root $(\cdot)^{1/2}$ is Lipschitz continuous with the Lipschitz constant depending on the minimum eigenvalue of $B$, and hence the map $f$ is locally Lipschitz.

It is not clear to me how to relax the positive definite condition of $B$ for the locally Lipschitz continuity.

John
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  • The answer is yes: take $C_t = \sqrt{A_tA_t^\top + BB^\top}$, where for a (symmetric) positive semidefinite matrix $M$, $\sqrt{M}$ is the unique positive semidefinite matrix satisfying $[\sqrt{M}]^2 = M$. – Ben Grossmann Apr 21 '22 at 15:39
  • @BenGrossmann Could you send me a reference about the property of the map? Why does it preserve the regularity of $A$? – John Apr 21 '22 at 19:36

1 Answers1

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The mapping $g:X\mapsto |X|:=(XX^T)^{1/2}$ that maps a square matrix to its polar part is Lipschitz continuous. This was first proved in Araki and Yamagami (1981), An inequality for the Hilbert-Schmidt norm, Comm. Math. Phys, 81:89-98 (see theorem 1 on p.89 and its proof on p.90). The authors proved that $\|\,|X|-|Y|\,\|_F\le\sqrt{2}\|X-Y\|_F$. The Lipschitz constant $\sqrt{2}$ is the best possible in the general case, but it can be improved to $1$ when $X$ and $Y$ are self-adjoint. See also section 5 of Rajendra Bhatia (1994), Matrix Factorizations and Their Perturbations, Linear Algebra and Its Applications, 197-198:245-276.

Since $\pmatrix{f(A)&0\\ 0&0}=\left|\pmatrix{A&B^{1/2}\\ 0&0}\right|$, your $f$ is also Lipschitz continuous with the same constant $\sqrt{2}$, but I don't know whether this constant can be improved or not.

user1551
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  • Thanks for your nice answer. Just to clarify, when you write $f(A)=\left|\begin{pmatrix} A & B^{1/2} \ 0 & 0\end{pmatrix}\right|$, do you mean they are equal in the norm? Strictly speaking, $f(A)$ is $n\times n$-dimensional, while the right-hand side is $(2n)\times (2n)$-dimensional (with all entries zero except the upper-left one). – John Jun 05 '22 at 00:56
  • If I understand the notation correctly, I think we should have the identity $\left|\pmatrix{A&B^{1/2}\ 0&0}\right|=\pmatrix{f(A)& 0\ 0&0}$. Please kindly let me know if I overlooked anything. – John Jun 05 '22 at 01:53
  • @John Surely it's a typo. – user1551 Jun 05 '22 at 05:28