Multivariate Normal Difference Distribution Since the distribution of a difference of two normally distributed variates X and Y with means and variances $(\mu_x,\sigma_x^2)$ and $(\mu_y,\sigma_y^2)$ respectively is given by another normal distribution with mean $\mu_x-\mu_y$ and variance $\sigma_x^2+\sigma_y^2$ http://mathworld.wolfram.com/NormalDifferenceDistribution.html. 
Assuming that the distributions are independent, can we find a distribution of a difference of two multivariate normally distributed variates?
 A: If two multivariate normal random variables $X\sim\mathcal N(\mu_X,\Sigma_X)$ and $Y\sim\mathcal N(\mu_Y,\Sigma_Y)$ are independent (and of the same dimension), then their sum is still normal, and you can sum the mean and variance directly:
$$X+Y\sim\mathcal N(\mu_X+\mu_Y,\Sigma_X+\Sigma_Y).$$
The easiest way to see this is with the expression of the characteristic function, as can be found on the Wikipedia article for example:
$$\phi_X(u)=\exp(iu'\mu_X-u'\Sigma_Xu).$$
Since the characteristic function of the sum of two independent random variables is the product of their respective characteristic functions, then
\begin{eqnarray}
\phi_{X+Y}(u)
&=&
\exp(iu'\mu_X-u'\Sigma_Xu)\exp(iu'\mu_Y-u'\Sigma_Yu)\\
&=&
\exp(iu'\mu_X-u'\Sigma_Xu+iu'\mu_Y-u'\Sigma_Yu)\\
&=&
\exp(iu'(\mu_X+\mu_Y)-u'(\Sigma_X+\Sigma_Y)u).
\end{eqnarray}
Now back to your original question, it suffices to see that $-Y$ has the same variance as $Y$, but with opposite mean to conclude:
$$-Y\sim \mathcal N(-\mu_Y,\Sigma_Y),$$
so indeed
$$X-Y\sim\mathcal N(\mu_X-\mu_Y,\Sigma_X+\Sigma_Y).$$
