# Positive definiteness of a matrix-valued function

This question is a repost from math.se, where I didn't receive an answer.

Are there simple conditions on an $$d \times d$$ matrix B under which $$f(t, s) = \begin{cases} \exp(-B |t - s|^\alpha), & t > s, \\[7pt] \exp(-B^\top |t - s|^\alpha), & t < s \end{cases}$$ with $$\alpha \in (0,2)$$ and $$t, \, s \in \Bbb R$$ is positive definite as a matrix-valued function (see def below)?

I know that it's true if $$B$$ is nonnegative definite (hence symmetric). But can these conditions be relaxed? Is it possible to show this for $$B$$ satisfying $$B + B^\top \geq 0$$ (eigenvalues have nonnegative real parts) without $$B$$ being symmetric (in particular, $$B$$ needs not be diagonalizable)?

Example. Consider a standard normal vector $$\mathbf{N}$$ and an antisymmetric matrix $$V$$. Define a Gaussian process $$\mathbf{X} (t) = \exp(t V) \, \mathbf{N}$$. Then $$\mathbb{E} \mathbf{X} (t) \, \mathbf{X}^\top (s) = \exp(V (t — s)) = \begin{cases} \exp(V |t — s|), & t > s, \\[7pt] \exp(V^\top |t — s|), & t < s \end{cases}$$ hence it is positive definite.

Is it true in general that if $$r(t, s)$$ is a positive definite matrix-valued function, then so is $$\exp(r(t, s))$$? This is obviously true if $$r$$ is scalar valued.

The reason for this particular question is that I am interested in the corresponding class of multivariate fractional Ornstein-Uhlenbeck processes.

P.S. By positive definiteness I mean $$\sum_{i, j} \mathbf{x}_i^\top f (t_i, t_j) \, \mathbf{x}_j \geq 0 \qquad \forall \, t_i \in \mathbb{R}, \quad \mathbf{x}_i \in \mathbb{R}^d.$$

First a remark: the usual and widely used definition for positive definiteness is what you wrote at the very end of your question $$\underline{\rm and}$$ the fact that the matrix $$F(t)=(f(t_i,t_j)_{1\le i, j\le n}$$ is non-singular. In other words, a matrix $$F=(f_{i,j})_{1\le i, j\le n}$$ is positive definite if $$\forall x\in \mathbb R^n, \quad x\not=0\Longrightarrow\langle F x, x\rangle>0. \tag{1}$$ Take now a $$n\times n$$ real-valued matrix satisfying (1); defining $$F_0=\frac12(F+F^T)\ \text{(a symmetric matrix)}, \quad F_1=\frac12(F-F^T) \ \text{(a skew-symmetric matrix)}.$$ Condition (1) means only that $$F_0$$ is a symmetric positive definite matrix and your question is in fact the following: let $$A$$ be a symmetric positive definite matrix, is it true that for any real-valued skew-symmetric matrix $$S$$, we have $$e^{A+S}$$ is positive definite?
The answer is negative: take for instance $$A=I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}, \quad S=π\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$ Since the matrices $$A,S$$ are commuting, we have $$e^{A+S}=e^Ae^S=\begin{pmatrix}e&0\\0&e\end{pmatrix} \begin{pmatrix}\cosπ&-\sin π\\\sin π&\cos π\end{pmatrix} =\begin{pmatrix}-e&0\\0&-e\end{pmatrix},$$ which is a negative-definite matrix.
• I don't see why this is true: "your question is in fact the following: let $A$ be a symmetric positive definite matrix, is it true that for any real-valued skew-symmetric matrix $S$, we have $e^{A+S}$ is positive definite?" Commented Mar 19, 2023 at 18:21
• @Iosif Pinelis The last question (about $r(t,s)$) is asking about $\exp r$ being positive-definite when $r$ is positive definite. Since $r$ can be written as the sum of a symmetric matrix and a skew-symmetric matrix, and moreover since the positiveness assumption does not concern the skew-symmetric part, you arrive at the question formulated in my answer ($A,S$ business). Commented Mar 20, 2023 at 9:55
• @Bazin, positive definiteness of $r(t, s)$ as a matrix-valued function includes symmetricity of in the sense that $r^\top(t,s) = r(s, t)$. Consider the example I added. Although admittedly I seriously doubt that $\exp(r(t,s))$ is positive definite, but for different reasons to what you've suggested. It doesn't even seem possible (at least to me) to show that the square $r^2(t, s)$ is positive definite. Commented Mar 20, 2023 at 11:18