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This reasoning is essentially that of Sycorax's answer, but no need to resort to that theorem:

Consider two points $x$ and $y$. For $\theta<0$, their Gram matrix is $$ \begin{bmatrix} k(x, x) & k(x, y) \\ k(x, y) & k(y, y) \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} $$ where $\alpha = k(x, y) = \exp\left( - \frac{\theta}{2} \lVert x - y \rVert^2 \right) = \exp\left( \tfrac12 \lvert{\theta}\rvert \lVert x - y \rVert^2 \right) > 1$, since the argument to $\exp$ is strictly positive.

The characteristic polynomial of this Gram matrix gives $(\lambda - 1)^2 - \alpha^2 = 0$, so that $\lvert \lambda - 1 \rvert = \alpha$, and the eigenvalues of this matrix are $1 + \alpha$ and $1 - \alpha$. Since $\alpha > 1$, that second eigenvalue is negative, and the kernel is not psd.

This reasoning is essentially that of Sycorax's answer, but no need to resort to that theorem:

Consider two distinct points $x$ and $y$. For $\theta<0$, their Gram matrix is $$ \begin{bmatrix} k(x, x) & k(x, y) \\ k(x, y) & k(y, y) \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} $$ where $\alpha = k(x, y) = \exp\left( - \frac{\theta}{2} \lVert x - y \rVert^2 \right) = \exp\left( \tfrac12 \lvert{\theta}\rvert \lVert x - y \rVert^2 \right) > 1$, since the argument to $\exp$ is strictly positive.

The characteristic polynomial of this Gram matrix gives $(\lambda - 1)^2 - \alpha^2 = 0$, so that $\lvert \lambda - 1 \rvert = \alpha$, and the eigenvalues of this matrix are $1 + \alpha$ and $1 - \alpha$. Since $\alpha > 1$, that second eigenvalue is negative, and the kernel is not psd.

This reasoning is essentially that of Sycorax's answer, but no need to resort to that theorem:

Consider two points $x$ and $y$. For $\theta<0$, their Gram matrix is $$ \begin{bmatrix} k(x, x) & k(x, y) \\ k(x, y) & k(y, y) \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} $$ where $\alpha = k(x, y) = \exp\left( - \frac{\theta}{2} \lVert x - y \rVert^2 \right) = \exp\left( \tfrac12 \lvert{\theta}\rvert \lVert x - y \rVert^2 \right) > 1$, since the argument to $\exp$ is strictly positive.

The eigenvalues of this matrix are $1 + \alpha$ and $1 - \alpha$: the characteristic polynomial is $(\lambda - 1)^2 - \alpha^2 = 0$, s that $\lvert \lambda - 1 \rvert = \alpha$. Since $\alpha > 1$, that second eigenvalue is negative, and the kernel is not psd.

This reasoning is essentially that of Sycorax's answer, but no need to resort to that theorem:

Consider two points $x$ and $y$. For $\theta<0$, their Gram matrix is $$ \begin{bmatrix} k(x, x) & k(x, y) \\ k(x, y) & k(y, y) \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} $$ where $\alpha = k(x, y) = \exp\left( - \frac{\theta}{2} \lVert x - y \rVert^2 \right) = \exp\left( \tfrac12 \lvert{\theta}\rvert \lVert x - y \rVert^2 \right) > 1$, since the argument to $\exp$ is strictly positive.

The characteristic polynomial of this Gram matrix gives $(\lambda - 1)^2 - \alpha^2 = 0$, so that $\lvert \lambda - 1 \rvert = \alpha$, and the eigenvalues of this matrix are $1 + \alpha$ and $1 - \alpha$. Since $\alpha > 1$, that second eigenvalue is negative, and the kernel is not psd.

This reasoning is essentially that of Sycorax's answer, but no need to resort to that theorem:

Consider two points $x$ and $y$. Their Gram matrix is $$ \begin{bmatrix} k(x, x) & k(x, y) \\ k(x, y) & k(y, y) \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} $$ where $\alpha = k(x, y) = \exp\left( - \frac{\theta}{2} \lVert x - y \rVert^2 \right) = \exp\left( \tfrac12 \lvert{\theta}\rvert \lVert x - y \rVert^2 \right) > 1$, since the argument to $\exp$ is strictly positive.

The eigenvalues of this matrix are $1 + \alpha$ and $1 - \alpha$: the characteristic polynomial is $(\lambda - 1)^2 - \alpha^2 = 0$, s that $\lvert \lambda - 1 \rvert = \alpha$. Since $\alpha > 1$, that second eigenvalue is negative, and the kernel is not psd.

This reasoning is essentially that of Sycorax's answer, but no need to resort to that theorem:

Consider two points $x$ and $y$. For $\theta<0$, their Gram matrix is $$ \begin{bmatrix} k(x, x) & k(x, y) \\ k(x, y) & k(y, y) \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} $$ where $\alpha = k(x, y) = \exp\left( - \frac{\theta}{2} \lVert x - y \rVert^2 \right) = \exp\left( \tfrac12 \lvert{\theta}\rvert \lVert x - y \rVert^2 \right) > 1$, since the argument to $\exp$ is strictly positive.

The eigenvalues of this matrix are $1 + \alpha$ and $1 - \alpha$: the characteristic polynomial is $(\lambda - 1)^2 - \alpha^2 = 0$, s that $\lvert \lambda - 1 \rvert = \alpha$. Since $\alpha > 1$, that second eigenvalue is negative, and the kernel is not psd.