Singular Value Decomposition on covariance matrix for multivariate normal distribution

Question

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!

Answer 1

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 $B = PL$.