Linear combination of random variables and independence Given a set of real-valued normally distributed random variables $X_1, \dots, X_n$, how do we show the following are equivalent?

*

*Any linear combination of $X_1, \dots, X_n$ is normally distributed

*There is a linear mapping $A$ such that the transformed random variables $AX_1, \dots, AX_n$ are independent

Motivation: I want to reconcile two definitions of jointly Gaussian random variables. I believe a set of scalar Gaussian rvs $\{X_i\}$ can be shown jointly Gaussian under two characterizations: 1) $\{X_i\}$ are independent under some linear transformation, or 2) all linear combinations of $\{X_i\}$ are Gaussian-distributed. I don't know why these are equivalent, or how to prove this property for a given set of rvs $\{X_i\}$.
Edit: if the covariance matrix is known, how does this help? I think one case where 1 and 2 don't hold is if the covariance is singular.
 A: I have to admit that I have slightly misread your question (I thought you would assume Gaussian random variables for 2). In that case the characteristic function would be the way to go).
Re-readng the question, I think there is problem in the formulation of 2) as it is stated right now. To see why suppose that $X_1, X_2, \dots, X_n$ are identically but dependent univariate random variables and let $A$ be a linear transformation such that $AX_1, AX_2,\dots, AX_n$ are independent. In the univarite case, $A$ is just a scalar. Assume $A\neq 0$. If $AX_1, AX_2, \dots, AX_n$ are independent, so is $A^{-1}AX_1, A^{-1}AX_2,\dots, A^{-1}AX_n$ which yields a contradiction. If $A=0$, the distribution is degenerate. Thus, an $A$ as claimed stated in 2) does not exist.
I assume you meant there exists a linear transformation $A$ such that $A\mathcal X$ is independent, where $\mathcal X = (X_1, X_2, \dots, X_n)$. But that still does not suffice to characterize the normal distribution (see, for example, https://stats.stackexchange.com/questions/74410/for-which-distributions-does-uncorrelatedness-imply-independence).

Edit: The following holds:
Theorem. Suppose that $X_1, X_2, \dots, X_n$ are real-valued random variables. Then $\mathcal X = (X_1, X_2,\dots, X_n)$ is normally distributed iff $A\mathcal X$ is normally distributed for every full row rank linear transformation $A$.
Proof. Let $A$ be a full row rank linear transformation and assume that $\mathcal X$ is normally distributed with mean $\mu$ and vcov matrix $\Sigma$. Since $A$ has full row rank, for each $s$ ($\in\mathbb R^k$) there is a $t$ such that $s = A't$. The characteristic function of $A\mathcal X$ is then given by $$\begin{align*}\varphi(t) &= \mathbb E[\exp(it'A\mathcal X)] \\ &=\mathbb E[\exp(i(t'A)\mathcal X)]\\&= \mathbb E[\exp(is'\mathcal X)] \\ &=\exp\left(is'\mu - \frac 12s'\Sigma s\right) \\ &= \exp\left(it'A\mu - \frac 12 t'A\Sigma A't\right),\end{align*}$$ which is the characteristic function of a $\mathcal N(A\mu, A\Sigma A')$-distributed random variable (note that the full row rank ensures that $A\Sigma A'$ is positive definite). Thus, $A\mathcal X$ is normally distributed as the characteristic function uniquely defines the distribution.
Now assume that $A\mathcal X$ is normally distributed for every full row rank linear transformation $A$. Consider the canonical basis vectors for $A$ to obtain that each component of $\mathcal X$ is normally distributed with finite variance. Using the Cauchy Schwartz inequality, it follows that the covariances are also finite. Thus, the vcov matrix $\Sigma$ of $\mathcal X$ as well as the mean $\mu$ exist. Furthermore, the mean and the variance of $A\mathcal X$ are given by $A\mu$ and $A\Sigma A'$, respectively. The characteristic function of $\mathcal X$ is given by
$$
\begin{align*}
    \varphi(s) &= \mathbb E[\exp(is'\mathcal X)] \\ 
     &= \mathbb E[\exp(it'A\mathcal X)] \\
     &= \mathbb E[\exp(it'(A\mathcal X))] \\ 
     &= \exp\left(it'(A\mu) - \frac 12 t'(A\Sigma A')t\right) \\ 
     &= \exp\left(i(A't)'\mu - \frac 12(A't)'\Sigma(A't)\right) \\ 
     &= \exp\left(is'\mu - \frac 12s'\Sigma s\right),
\end{align*}$$ using again the fact that $A$ has full row rank. This is the characteristic function of a $\mathcal N(\mu,\Sigma)$-distributed random variable, and the characteristic function uniquely defines the distribution. $\square$
A special case is, for example, $A = a'$, for some $a\in\mathbb R^n$, $a\neq 0$. Another example is $A = \Sigma^{-\frac 12}$, which yields a linear combination of $\mathcal X$ that renders the components independent by de-correlation.
