With $UD V^T$ is the SVD of $X$, is $X\mapsto UV^T$ Lipschitz over the set of matrices with full rank and smallest singular value at least $a>0$?

Question

Let $m\le n$. For any matrix $X\in R^{m\times n}$ with full-rank $m$, define the usual SVD $X = UD V^T$.

Define the open set $M_a = \{X\in R^{m\times n}: \lambda_{\min}(XX^T)> a\}$ where $\lambda_{\min}$ is the smallest eigenvalue.

Is it true that $X\mapsto U V^T$ is Lipschitz continuous on $M_a$ with respect to the operator norm or Frobenius norm, and what is the correct dependence of the dimension(s) on the Lipschitz constant?

Answer 1

For the Frobenius norm, the answer is positive using:

Andersson, Fredrik, Marcus Carlsson, and Karl-Mikael Perfekt. "Operator-Lipschitz estimates for the singular value functional calculus." Proceedings of the American Mathematical Society 144.5 (2016): 1867-1875.

Indeed, define $f: R_{\ge0} \to R$ by $f(x) = 1$ if $x>a$ and $f(x)=x/a$ if $x\in[0,a]$. The Lipschitz constant of $f$ is $1/a$. By Theorem 1.1 in the above reference in the real-valued case, if two matrices $A,B$ have SVD $A=UDV^T$ and $B=\hat U \hat D \hat V^T$ then $$ \|U f(D) V^T - \hat U f(\hat D) \hat V^T\|_F \le (1/a) \|A - B\|_F. $$ If $A,B\in M_a$ then $f(D)=I=F(\hat D)$ so this answers the question in the Frobenius case.