2
$\begingroup$

I am trying to figure out whether the Cholesky decomposition is uniformly continuous within a set of matrices with bounded entries. Specifically, I would like to derive a modulus of continuity for the Cholesky decomposition with respect to the spectral matrix norm, i.e., a monotonically increasing function $\omega: \mathbb{R}_+ \rightarrow \mathbb{R}_+$, such that $\Vert\mathbf{L}_A-\mathbf{L}_B \Vert \leq \omega(\Vert\mathbf{A}-\mathbf{B} \Vert)$ holds for all symmetric positive definite matrices $\mathbf{A} = \mathbf{L}_A \mathbf{L}_A^{\top}$ and $\mathbf{B} = \mathbf{L}_B \mathbf{L}_B^{\top}$ with $[\mathbf{A}]_{ij}, [\mathbf{B}]_{ij} \leq C$ for all $i,j \leq n$ and some $0< C < \infty$. Here $\Vert \cdot \Vert$ denotes the induced (spectral) norm $\Vert \mathbf{A} \Vert = \max_{\Vert\mathbf{x}\Vert=1} \Vert\mathbf{A}\mathbf{x} \Vert$.

I know that a similar bound holds for the positive-definite square root of a matrix, i.e., $\Vert \sqrt{\mathbf{A}} - \sqrt{\mathbf{B}} \Vert \leq C \Vert \mathbf{A} - \mathbf{B}\Vert^{\frac{1}{2}}$ for a constant C that potentially depends on the dimension of the matrices [1]. I am also aware of bounds on the derivative of the Cholesky decomposition mapping [2], which allows us to obtain bounds that depend on the condition number of the matrix $\mathbf{A}$. However, I require a bound that does not include the condition number, nor the norm of the inverse of $\mathbf{A}$ or $\mathbf{B}$ .

[1] - Rajendra Bhatia, Matrix Analysis, Theorem X.l.l

[2] - Rajendra Bhatia, Matrix Analysis, Problem X.5.9

$\endgroup$
2
  • 2
    $\begingroup$ Would you mind changing the link? I can't access it. Generally it's probably wise to include some information to identify the contents, e.g. the usual bibliographical data if this is a paper. $\endgroup$
    – Leo Moos
    Apr 15, 2021 at 18:02
  • $\begingroup$ Thanks for the hint. I have included the sources in the question, but do not know of a publicly available link (the one I provided can only be accessed by me, I only became aware of this after you mentioned it). $\endgroup$
    – Heinrich A
    Apr 16, 2021 at 10:13

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.