TITLE: When linear combinations of independent random variables are still independent?
QUESTION [11 upvotes]: We have a random vector $X=(X_j)_{j=1,...,n}$, whose components $X_j$ are mutually independent. We build a new random vector $Y=A X+b$, with $Y=(Y_k)_{k=1,...,m}$, $A=(a_{k,j})_{k=1,...,m;j=1,...,n}$, $b=(b_k)_{k=1,...,m}$.
When the components of $Y$ are still mutually independent random variables? How to prove the result? I suppose this can be related with the rank of $A$ but I'm not sure and I cannot find a way to prove if I don't make some special assumption about the law of $X$.
I know $U$ and $V$ are independent iff $f(U)$ and $g(V)$ are independent for each and every pair of measurable function $f$ and $g$. This can help, I think, when, for example, $Y_1=X_1+X_2$ and $Y_2=X_3+X_4$. 
But when does $Y_1 = X_1 + X_2$ and $Y_2 = X_1 - X_2$? Or when does $Y_1=X_1 + X_2$ and $Y_2=X_5$?  
None of the texts I searched discuss seriously this topic, but leave it in a certain way to "intuition"?
Thanks

REPLY [3 votes]: The Skitovich-Darmois theorem A (Skitovich(1953), Darmois(1953), see also
A. Kagan, Yu. Linnik, and C.R. Rao (1973, Ch.3)). Let $\xi_j$, where $j=1,
2,\dots, n,$ and $n\geq 2$, be independent random variables. Let
  $\alpha_j, \beta_j$  be nonzero constants. If the linear statistics
$L_1=\alpha_1\xi_1+\cdots+\alpha_n\xi_n$ and
$L_2=\beta_1\xi_1+\cdots+\beta_n\xi_n$ are independent, then all
random variables $\xi_j$ are Gaussian.