# Operator norm of difference of matrix decompositions

This question is in part related to a question that I have already posed.

Say I have two symmetric positive definite matrices and their respective Cholesky decompositions $$\mathbf{A} = \mathbf{L}_A \mathbf{L}_A^{\top}$$ and $$\mathbf{B} = \mathbf{L}_B \mathbf{L}_B^{\top}$$, where $$\mathbf{L}_A$$ and $$\mathbf{L}_B$$ have positive diagonal entries. Furthermore, let $$\sqrt{\mathbf{A}}$$ and $$\sqrt{\mathbf{B}}$$ denote the unique symmetric positive definite square roots of $$\mathbf{A}$$ and $$\mathbf{B}$$, respectively, i.e., $$\sqrt{\mathbf{A}}^{\top} = \sqrt{\mathbf{A}}$$ and $$\sqrt{\mathbf{A}}\sqrt{\mathbf{A}} = \mathbf{A}$$.

I would like to know whether there is an inequality relating the operator norms of $$\mathbf{L}_A - \mathbf{L}_B$$ and $$\sqrt{\mathbf{A}}- \sqrt{\mathbf{B}}$$ up to a positive constant $$C>0$$, i.e., $$$$\max_{\mathbf{x}} \frac{\Vert (\mathbf{L}_A - \mathbf{L}_B)\mathbf{x}\Vert}{\Vert\mathbf{x}\Vert} \leq C \max_{\mathbf{x}} \frac{\Vert(\sqrt{\mathbf{A}} - \sqrt{\mathbf{B}})\mathbf{x}\Vert}{\Vert\mathbf{x}\Vert}.$$$$

Here $$\Vert \cdot \Vert$$ denotes the Euclidean norm.

Edit: As pointed out by Denis Serre, the inequality does not hold in general. However, I can also include the assumption that the entries of $$\mathbf{A}$$ and $$\mathbf{B}$$ are bounded (and therefore also their square-roots and Cholesky decompositions). From Denis' answer, it seems to me that Lipschitz continuity from $$\sqrt{\mathbf{A}}$$ to $$\mathbf{L}_A$$ would hold, since the derivative of $$\mathbf{L}_A$$ with respect to the entries of $$\sqrt{\mathbf{A}}$$ cannot become unbounded.

• Thanks for the suggestion. Equality does not hold in general. I have realized that I actually require inequality instead of equality, as specified in the new edit. Apr 19, 2021 at 11:56

The answer is negative, and this happens as soon as $$n=2$$. The question is whether the composition $$X\mapsto L:=L_{X^2}$$ is globally Lipschitz over $${\bf SPD}_n$$. Let $$x_j\in{\mathbb R}^n$$ denote the $$j$$th column of $$X$$. Then we have the following formulae $$\ell_{11}=\|x_1\|,\quad \ell_{j1}=\frac{\langle x_1,x_j\rangle}{\|x_1\|}\,,$$ and so on, the expression getting more and more complicated as the column index increases.
The first entry $$l_{11}$$ is obviously globally Lipschitz. But this is not the case for the next ones. We have $$d\ell_{j1}=\frac{x_1}{\|x_1\|}\cdot dx_j+x_j\cdot d\frac{x_1}{\|x_1\|}.$$ Because the entries $$x_{kj}$$ are independent from $$x_1$$ if $$k\ge2$$, the global Lipschitz property can hold only if $$d\frac{x_1}{\|x_1\|}\parallel \vec e_1,$$ which is not true in general.
Edit. The obstacle discussed above occurs at infinity, because there is no a priori bound of $$x_{22}$$, hence $$d\ell_{21}$$ is not uniformly bounded as $$x_{22}\rightarrow+\infty$$. But because $$X\mapsto L$$ is homogeneous of degree $$1$$, this has the counterpart that the Lipschitz property also fails near the origin. Actually the norm of $$d\frac{x_1}{\|x_1\|}=\frac1{\|x_1\|}\left(I_n-\frac{x_1\otimes x_1}{\|x_1\|^2}\right)dx_1$$ is $$1/\|x_1\|$$, which can be arbitrarily large. Thus even if we impose an a priori bound on $$X$$, the map $$X\mapsto L$$ is not Lipschitzian.
• Thanks very much for the helpful answer! I was wondering whether the composition would be globally Lipschitz if I werer to introduce the assumption that the columns entries of the matrices $\mathbf{A}$, $\mathbf{B}$ (and therefore also their square roots and Cholesky decompositions) are bounded. Apr 19, 2021 at 15:07
• @HeinrichA. See my edit: even if we impose an a priori bound, the map $X\mapsto L$ is not Lipschitz. Apr 20, 2021 at 6:44