6
$\begingroup$

Let $\Bbb{S}_{++}^n$ be the $\frac{n(n+1)}{2}$-dimensional Riemannian manifold of the symmetric positive definite (SPD) $n\times n$ real matrices.

The Log-Euclidean distance between two points of $\Bbb{S}_{++}^n$, i.e. between two SPD matrices $A,B\in\Bbb{S}_{++}^n$, is given by $$ d(A,B)=\lVert\log(A)-\log(B)\rVert_{F}, $$ where $\|\cdot\|_{F}$ denotes the regular Frobenius norm.

We want to prove that $d\colon\Bbb{S}_{++}^n\times\Bbb{S}_{++}^n\to\Bbb{R}$ is a conditionally negative definite function, as defined below.

For a topological space $\mathcal{X}$, the function $f\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$ is called conditionally negative definite if for any $m\in\Bbb{N}$, $x_1,\ldots,x_m\in\mathcal{X}$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true $$ \sum_{i,j=1}^{m}c_ic_jf(x_i,x_j)\leq0. $$

In our case, we want to prove that for any $m\in\Bbb{N}$, $X_1,\ldots,X_m\in\Bbb{S}_{++}^n$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true $$ \sum_{i,j=1}^{m}c_ic_j\lVert\log(X_i)-\log(X_j)\rVert_{F}\leq0. $$ Thank you very much in advance!

Edit: In case of the squared Frobenius norm, I think it would be proven more easily, but I still need some help...

$\endgroup$
5
  • $\begingroup$ By substituting $Y_i:=\log(X_i)$ you are left with proving the same statement for euclidean spaces, which I guess is true. $\log$ is a bijection from the space of SPD-matricses to the space of symmetric matrices. $\endgroup$
    – user35593
    Jul 3, 2015 at 17:16
  • $\begingroup$ First, thanks for your comment. Well, I actually tried that, even in the simpler case of $\mathbf{x}_i\in\Bbb{R}^n$: $\|\mathbf{x}_i-\mathbf{x}_j\|_2$. The problem here is the square root. I cannot figure that out; apparently, I cannot use the triangle inequality... If I had the squared norm, i.e. $\|\mathbf{x}_i-\mathbf{x}_j\|_2^2$ thinks become much easier. $\endgroup$ Jul 3, 2015 at 17:30
  • $\begingroup$ Why does this have the at.algebraic-topology tag? $\endgroup$ Jul 6, 2015 at 10:58
  • $\begingroup$ @SeanTilson, apologies, that's a mistake. I edit it. $\endgroup$ Jul 6, 2015 at 10:59
  • 1
    $\begingroup$ There is no need for an apology, I was just curious. :) $\endgroup$ Jul 6, 2015 at 11:16

1 Answer 1

4
$\begingroup$

First, recall that if $\psi$ is negative definite, then $\exp(-\gamma \psi)$ is positive definite for all $\gamma >0$. Now, from Corollary 2.10 of Harmonic analysis on Semigroups (Berg, Christensen, Ressel) we also know that if $\psi$ is negative definite, then $\psi^\alpha$ is also negative definite for $0 < \alpha < 1$ as long as $\psi(x,x) \ge 0$.

Consider now some map $f: \mathcal{X} \to \mathbb{R}^n$, and set $\psi(x,y) = \|f(x)-f(y)\|^2$. Clearly, $\psi$ is negative definite.

In particular, choosing $\mathcal{X} = \mathbb{S}^n_{++}$, $f \equiv \log$, and $\alpha=1/2$, from the above corollary it follows that $\|\log X - \log Y\|_\text{F}$ is negative definite.

$\endgroup$
7
  • $\begingroup$ Thank you very much for your answer! Let me clarify some issues before I know I understand the answer. First, in fact, I need that result in order to prove the positive definiteness of a kernel of the form you mentioned ($\exp(-\gamma d(x,y))$). Now, if I understand your answer, you don't use that fact. The rest of the answer is quite clear to me, I think. So, I prove the negative-definiteness of the squared distance, then I use the corollary you mentioned, am I right? $\endgroup$ Jul 3, 2015 at 19:05
  • $\begingroup$ I just wrote that as a "definition" --- that equivalence gets used when proving the cited corollary, hence I left it in there. $\endgroup$
    – Suvrit
    Jul 3, 2015 at 19:15
  • $\begingroup$ Nice, a last question please. Is there any other source I could use for citing that Corollary? Thank you very much again! $\endgroup$ Jul 3, 2015 at 19:17
  • $\begingroup$ You can just include a proof of the corollary (while still citing that book). The proof follows from the following integral representation: $$\psi^\alpha = \frac{\alpha}{\Gamma(1-\alpha)}\int_0^\infty (1-\exp(-\gamma \psi))\frac{d\gamma}{\gamma^{\alpha+1}},$$ which is valid for $\psi \ge 0$ (the proof follows since $1-\exp(-\gamma \psi)$ is negative definite due to $\exp(-\gamma \psi)$ being pd, and all other terms are nonnegative). $\endgroup$
    – Suvrit
    Jul 3, 2015 at 19:23
  • $\begingroup$ However, I don't want to use the fact that $\exp(-\gamma\psi)$ is positive-definite, since this is exactly I want to prove by the proving the negative-definiteness of $\psi$. Apologies for my ignorance, I just got a bit confused... $\endgroup$ Jul 3, 2015 at 19:26

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.