A Gaussian measure $\mu$ on $\mathbb{R}^{N}$ is said to be Gaussian if its Fourier transform has the form: $$\hat{\mu}(y) := \int_{\mathbb{R}^{N}}e^{i\langle y,x\rangle}d\mu(x) = e^{i\langle a,y\rangle - \frac{1}{2}\langle Ky,y\rangle}$$
where $a \in \mathbb{R}^{N}$ and $K$ is a nonnegative symmetric matrix. I want to prove that if $\mu$ is Gaussian, then $\mu$ has density if, and only if $K$ is nondegenerate. I already proved that if $K$ is nondegenerate, then $\mu$ has density, but I'm stuck at the converse: if $\mu$ has density, then $K$ needs to be nondegenerate. Any hints, please?
The Riemann-Lebesgue lemma tells us that the Fourier transform of an $L^{1}$ function vanishes at infinity. But if $K$ is not positive definite there is some $y$ for which $\left\langle K y, y\right\rangle\le 0$. So $\mathcal{F}\mu$ will not be heading to zero in that direction.