A kernel is called positive definite (p.d) if its Gram matrix is p.d., i.e. all eigenvalues of the Gram matrix are positive for all possible input vectors in the feature space. My understanding of conditionally positive definite (c.p.d) kernels is that the Gram matrix has to be p.d. not for all but only for a subset of the input vectors from the feature space.

If this definition of conditionally positive definiteness is correct then how do I prove that a kernel is c.p.d.? Would it suffice to show that the Gram Matrix is p.d. for some arbitrary subset of input vectors?

  • $\begingroup$ The article by Charles A. Michelli may help: "Interpolation of scattered data : Distance matrices and conditionnally positive definite functions", in Constructive Approximation (1986) 2: 11-22 $\endgroup$ Mar 26 at 14:50

1 Answer 1


If I'm reading it correctly, your definition of cpd would include any kernel where $k(x, x) > 0$ for any point $x$, by just considering the subset $\{ x \}$.

One way to phrase a matrix being positive definite is that $c^T A c > 0$, for all vectors $c$; a kernel is positive definite if that holds for all Gram matrices $A$. The conditioning in cpd refers to reducing the "for all vectors $c$" part, not the "for all Gram matrices $A$" part. Specifically, the most common definition is reducing it to "for all vectors $c$ with $1^T c = 0$.

These are useful because, just as pd kernels correspond to inner products in some Hilbert space, cpd kernels are closely connected to distances in some Hilbert space. SVMs and kernel PCA both work with any cpd kernel, though a particular optimization algorithm might not.

Here's the full definition from Schölkopf and Smola's Learning with Kernels (2002). Here $\mathbb K$ refers to either $\mathbb R$ or $\mathbb C$.

Definition 2.20 (Conditionally Positive Definite Matrix) A symmetric $m \times m$ matrix $K$ ($m \ge 2$) taking values in $\mathbb K$ and satisfying

$$ \sum_{i, j = 1}^m c_i \bar{c}_j K_{ij} \ge 0 \text{ for all } c_i \in \mathbb K, \text{ with } \sum_{i=1}^m c_i = 0$$

is called conditionally positive definite (cpd).

Definition 2.21 (Conditionally Positive Definite Kernel) Let $\mathcal X$ be a nonempty set. A function $k : \mathcal X \times \mathcal X \to \mathbb K$ which for all $m \ge 2, x_1, \dots, x_m \in \mathcal X$ gives rise to a conditionally positive definite Gram matrix is called a conditionally positive definite (cpd) kernel.

I don't know a simple eigenvalue-based definition.

To prove a kernel is cpd, you can either prove it directly from the definition above (usually difficult), or you can try to build it out of other cpd kernels using various properties that preserve cpd-ness. Here are some useful connections to pd kernels from that book:

Proposition 2.22 (Constructing PD Kernels from CPD Kernels [42, Lemma 2.1, p. 74]) Let $x_0 \in \mathcal X$, and let $k$ be a symmetric kernel on $\mathcal X \times \mathcal X$. Then $$\tilde k(x, x') := \tfrac12 \left( k(x, x') - k(x, x_0) - k(x_0, x') + k(x_0, x_0) \right)$$ is positive definite if and only if $k$ is conditionally positive definite.

[42] is Berg, Christensen, and Ressel's Harmonic Analysis on Semigroups (1984).

Proposition 2.28 (Connection PD – CPD [465, Thm. 1, p. 527]) A kernel $k$ is conditionally positive definite if and only if $\exp(t k)$ is positive definite for all $t > 0$.

[465] is the classic paper Metric spaces and positive definite functions of Schoenberg (Transactions of the American Mathematical Society 44:522-536, 1938).

  • $\begingroup$ Yes Definition 2.21 was actually what led to my understanding of (c.p.d) above... "A function $k : \mathcal X \times \mathcal X \to \mathbb K$ which for all $m \ge 2, x_1, \dots, x_m \in \mathcal X$ gives rise to a conditionally positive definite Gram matrix is called a conditionally positive definite (cpd) kernel" If I set m to 2 then there are only two $x_i$ left and therefore the Gram Matrix is only a $2 \times 2$ symmetric matrix $\endgroup$
    – nico1510
    May 6, 2015 at 8:51
  • $\begingroup$ It has to be true for all $m$ and sets of feature vectors $\{x_i\}$. An equivalent statement holds for positive definite kernels, except there we require it to hold for all weight vectors $c$ rather than just those with sum 0. $\endgroup$
    – Danica
    May 6, 2015 at 8:59
  • $\begingroup$ To be more concrete I'm using a distance measure which I want to substitute for e.g. the squared Eucledian Distance. My distance measure is nonnegative, symmetric, zero-diagonal like it is demanded in this paper. However it is no distance metric and on page 3 of the linked article the authors state that Corollary 1 (Non-Metricity Prevents Definiteness) and that only for certain parameter values the kernel is (c.p.d) $\endgroup$
    – nico1510
    May 6, 2015 at 9:02
  • $\begingroup$ With this information is it safe to assume that the resulting kernel is if at all for certain parameter values (c.p.d) ? $\endgroup$
    – nico1510
    May 6, 2015 at 9:05
  • 1
    $\begingroup$ Their Corollary 1 shows that there exist parameter values for which those kernels are not (c)pd. There do exist some parameter values for which they are: trivially, $k(x,y) = -d(x, y)^0 = -1$ is cpd. Whether there exist other (nontrivial) parameter values for which any of those kernels are (c)pd remains an open question, dependent on the distance itself. To answer it requires looking at more than just size $m=2$ kernel matrices; it probably requires something clever, or something in the definition of the distance. $\endgroup$
    – Danica
    May 6, 2015 at 9:10

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