Does a characteristic function in the format $e^{v^{T}Kv}$ implies a random vector with a Gaussian distribution?

Question

I know that if I have a Gaussian random vector $X$ with mean $m_X$=$[0]$, then I have a characteristic function $M_x(v)=e^{-\dfrac{1}{2}v^{T}Kv}$. In the other hand, if I have a characteristic function $M_x(v)=e^{-\dfrac{1}{2}v^{T}Kv}$ and a known mean $m_X$=$[0]$, then I can say $X$ is a Gaussian random vector. But if I only have the characteristic function $M_x(v)=e^{-\dfrac{1}{2}v^{T}Kv}$, can I say both $X$ is a Gaussian vector and it has mean $m_X$=$[0]$?