If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of the random vector also gaussian? Suppose I have a $k$-dimensional random vector $X\sim N(\mu,\Sigma)$, where $\mu$ is the mean vector and $\Sigma$ the covariance matrix.
Is the $i$-th element of $X$ also distributed as a gaussian with $X_i \sim N(\mu_i,\Sigma_{ii})$?
 A: A random vector $X$ have the multinormal distribution if all linear combinations are normally distributed. Just take the linear combination with coefficient vector $e_i = (0,0,\dots, 1,0,\dots,0)$ where the one is in place $i$. Then $e_i^T X=X_i$, the $i$th component, hence that is normally distributed.
For the additional question in the comments about mean and variance of $X_i$.
For the expectation (mean) of a linear combination we have 
$$
\DeclareMathOperator{\E}{E}  \E \sum_i a_i X_i = \sum_i a_i \E X_i = \sum_i a_i \mu_i
$$
and then you an conclude using the above! 
For the variance of a linear combination, use that the covariance of linear combinations are given by
$$
\DeclareMathOperator{\Cov}{Cov} 
\Cov(\sum_i a_i X_i, \sum_j b_j Y_j) = \sum_i \sum_j a_i b_j \Cov(X_i,Y_j)
$$
then that
$$
\DeclareMathOperator{\Var}{Var}
\Var(\sum_i a_i X_i)=\Cov(\sum_i a_i X_i, \sum_j a_j X_j)=
\sum_i \sum_j a_i a_j \Cov(X_i, X_j)
$$
finally putting the coefficient vector $e_i$ into this formulas.
So, we are using that expectation is a linear operator, while covariance is a bilinear operator.
A: kjetil's answer (+1) addresses the more general case of linear combinations of components of your multivariate normal distribution.
Your specific case concerns the so-called marginals of the multivariate normal distribution. And yes, the marginals of a multivariate normal distribution are again normal (multivariate normal, if you look at higher-dimensional marginals than your one-dimensional ones).
Interestingly, and importantly, the converse does not hold. A multivariate distribution can have normal marginals, but be non-multivariate normal. This is a standard homework problem. See, e.g., Kowalski (1973, The American Statistician), or here.
This indicates the importance of the "all" in kjetil's answer - it's not enough for only some linear combinations of marginals to be normal, all linear combinations need to be normal for the whole vector to be multivariate normal.
