Gaussian distribution and independence

Question

Suppose you have Z=$(Z_1,...,Z_n)^T$, which has the standard Gaussian distribution on $\Bbb R^n$, and the $m$ x $n$ matrix A. Let W=$(W_1,...,W_m)^T$ be defined by W=AZ. Since Z is standard Gaussian, it means that each $Z_i$ are independent. Since W is a function of Z, W has a multivariate normal distribution. However, will the $W_i$, i=1,2,...,m be independent i.e will the variance covariance matrix of W, which is $AA^T$, be a diagonal matrix?

Answer 1

No. $AA^\intercal$ has no reason to be diagonal e.g. $A=\begin{pmatrix}1&1\\1&0\end{pmatrix}$, then $AA^\intercal=\begin{pmatrix}2&1\\1&1\end{pmatrix}$.

You should know that any Gaussian vector is of the form $\mu+\sigma Z$, where $Z$ is the standard Gaussian vector. It is certainly false in general that the components of a Gaussian vector must be independent.