Let $X$ be an $n$-dimensional Gaussian random vector with mean $0$ and covariance $AA^T$. Clearly, if $Z$ is an $n$-dimensional Gaussian random vector with mean $0$ and covariance $I$ (the identity matrix), then $X$ and $AZ$ have the same distribution.
However, can I say that there exists $Z$ with mean $0$ and covariance $I$ such that $X = AZ$? If $A$ were invertible, I can let $Z = A^{-1}X$. What about for the general case where $A$ is an aribtrary matrix.