Can linear combinations of any Gaussian random variables be independent? Suppose that $X=[X_1\; X_2]^t$ is Gaussian vector. My question is whether $U=a_1X_1+a_2X_2$ and $V=b_1X_1+b_2X_2$, where $a_1b_2-a_2b_1\ne 0$, can be independent Gaussian random variables, if $a_1a_2b_1b_2 \neq 0$.   
 A: If $X_1$ and $X_2$ are independent normal random variables, then at least after normalizing the random variables to have equal variance (for simplicity) you can choose any two linear combinations $a_1 X_1 + a_2 X_2$ and $b_1 X_1 + b_2 X_2$ such that $a$ and $b$ are orthogonal, i.e. $a_1b_1 + a_2b_2 = 0$, and the resulting linear combinations will be independent. You can check this by computing the covariance between the two linear combinations, noting that $X_1$ and $X_2$ are independent and have equal variance. For example, you can choose $a_1 = 1, a_2 = 1, b_1 = 1, b_2 = -1$. This satisfies your conditions on the coefficients.
A: Let $X=[X_1\; X_2]^t$ be a jointly Gaussian vector and $K_X=E\{XX^t\}$ be its covariance matrix.  Since any covariance matrix is a positive semi-definite, it can be decomposed as $$K_X=W\Lambda W^t,$$
where $W=[w_1|w_2]$ is a unitary matrix and $\Lambda $ is a diagonal matrix. Then, since $w_1$ and $w_2$ are orthogonal, i.e., $w_1^tw_2=0$, we can prove that $U=w_1^tX$ and  $U=w_2^tX$ are uncorrelated because 
$$E\{UV\}=E\{UV^t\}=E\{w_1^tXX^tw_2\}=w_1^tE\{XX^t\}w_2=w_1^tK_Xw_2=\Lambda.$$ 
On the other hand, we know that if two jointly Gaussian random variables are uncorrelated they are independent. Therefore, if $X_1$ and $ X_2$ are jointly Gaussian, one can choose  $[a_1 \; a_2]=w_1^t$ and $[b_1 \; b_2]=w_2^t$. Note that the elements of $w_1$ and $w_2$ can be non-zero in general.

Therefore the answer to the question is YES!

