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., \begin{equation} \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}. \end{equation}

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.

  • $\begingroup$ 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. $\endgroup$
    – Heinrich A
    Apr 19, 2021 at 11:56

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Heinrich A
    Apr 19, 2021 at 15:07
  • 1
    $\begingroup$ @HeinrichA. See my edit: even if we impose an a priori bound, the map $X\mapsto L$ is not Lipschitz. $\endgroup$ Apr 20, 2021 at 6:44

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