Singular Value Decomposition on covariance matrix for multivariate normal distribution Suppose $x$ is MVN($0_n$, $I_n$), how to find $a$ and $B$ such that $a+Bx$ is MVN($\mu$, $\Sigma$)?  
Here is what I try: 
$a$ is easy to find: $$a = \mu$$   
for B: 
$$Cov(Bx) = BI_nB^T = \Sigma$$   
The problem is to find matrix $B$ using SVD.
Anyone help with how to perform the SVD here? 
Thanks!

Let $\Sigma = UDU'$ is the SVD decomposition of a positive definite matrix $\Sigma$. Then $a = \mu$ and $B = U D^{1/2}$. 
When $\Sigma$ is only semi-positive definite, then $\Sigma = UDV'$, possibly with  $V \neq U$, but can still take $B = U D^{1/2}$.
Alternatively you can perform the pivoted Cholesky decomposition of $\Sigma$:
$$\Sigma = (PL) \times (PL)'$$
where $L$ is lower triangular, and $P$ is a permutation matrix. Then $B = PL$.
