We know that if $X \sim N_p(\mu, \Sigma)$ then $a'X \sim N(a'\mu,a'\Sigma a)$ for and $a \in \mathbb{R}_p$. What I need to know is if the converse of this is also true. Can this be proved? Would appreciate any assistance. Thanks
Edit: Now I know this is true from what I have copied from Johnson's and Wichern's Applied multivariate statistical analysis. Only thing required now is the proof.
Result 4.2 If $X$ is distributed normally as $N_p(\mu, \Sigma)$ then any linear combination of variables $a'X = a_1X_1 + a_2X_2 + \cdots + a_pX_p$ is distributed as $N(a'\mu, a'\Sigma a)$. Also, if $a'X$ is distributed as $N(a'\mu, a'\Sigma a)$ for every $a$, then $X$ must be $N_p(\mu, \Sigma)$.
(As Did commented "Characteristic functions Characterize Distributions".) Assume that $a'X\sim N(a'\mu,a'\Sigma a)$ for every $a\in\mathbb{R}^p$. (Here $a'$ represents the transpose of the vector $a$). This is equivalent to the fact that the characteristic function $\Phi$ of $a'X$ is given by $$ \Phi(t)=\mathbb{E}\left(e^{ita'X}\right)=\exp\left(ita'\mu-\frac{1}{2} t^2a'\Sigma a\right) $$ In particular, for $t=1$, and every $a\in\mathbb{R}^p$ we have $$ \mathbb{E}\left(e^{i a'X}\right)=\exp\left(i a'\mu-\frac{1}{2} a'\Sigma a\right) $$ But this means that $X\sim N(\mu,\Sigma)$.$\qquad\square$