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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 $a_1a_2b_1b_2 \neq 0$.

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    $\begingroup$ I believe the answer is yes. Somebody posted an example in a question a couple days ago. I believe it was $X,Y,Z$ are independent standard normals, then $Z-Y+X$ and $X+Y$ are independent. You can prove it easily using the bilinearity property of covariance. That's not exactly what you are looking for because it involves a combination of three instead of two, but I imagine an example using two is not hard to come by $\endgroup$ – Gregory Grant Jun 5 '15 at 20:34
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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!

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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.

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