Showing $\sum a_iX_i$ and $\sum b_iX_i$ are independent iff $\sum a_ib_i=0$ where $X_i's$ are i.i.d $N(\mu,\sigma^2)$ 
Let $X_1, X_2,\ldots, X_n$ be i.i.d with the distribution $N(\mu, \sigma^2)$.
  Prove $Y =\sum_{i=1}^{n} a_iX_i$ and $Z =\sum_{i=1}^{n} b_iX_i$ are independent iff $\sum a_ib_i=0$.  

I have proved it by Basu's theorem.  I am searching for another methods.
For example, if I could prove that $(Y,Z)\sim N(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,\rho)$, it seems easier to show their independence.
 A: In general if $X_1,\dots,X_n$ are independent and all have normal distribution then random vector $\mathbf X:=(X_1,\dots,X_n)^T$ has normal distribution.
To get light on this realize that there is a common PDF that is the product of the PDF's of the $X_i$.
Then also in general random vector $A\mathbf X$ has normal distribution where $A$ is an $m\times n$ matrix.
Now note that $(Y,Z)^T$ can be written as $A\mathbf X$ where $A$ is an $2\times n$ matrix.
First row of $A$ is $(a_1,\dots, a_n)$ and second row of $A$ is $(b_1,\dots, b_n)$.

Here $Y$ and $Z$ have a joint normal distribution and in that case they are independent if they are uncorrelated. 
So it suffices to prove that their covariance equals $0$. 
Using bilinearity of $\mathsf{Cov}$ we find:$$\mathsf{Cov}(Y,Z)=\sum_{i=1}^n\sum_{j=1}^na_ib_j\mathsf{Cov}(X_i,X_j)=\sigma^2\sum_{i=1}^na_ib_i$$
So that $$\mathsf{Cov}(Y,Z)=0\iff\sum_{i=1}^na_ib_i=0$$
A: Since your variables are jointly Gaussian, then the independence of $Y$ and $Z$ is given by their decoration. Observe that
\begin{align*}
\text{Cov}(Y,Z) &= \text{Cov}(\sum_{i=1}^n a_i X_i, \sum_{j=1}^n b_j X_j)\\
&=\sum_{i=1}^n\sum_{j=1}^n a_i b_j^\ast\times \text{Cov}( X_i, X_j)\\
&=\sum_{i=1}^n\sum_{j=1}^n a_i b_j^\ast\times \sigma^2\mathbf{1} (i=j)\\
&=\sigma^2 \sum_{i=1}^n a_i b_j^\ast
\end{align*}
So $X$ and $Y$ are independents if and only if $\text{Cov}(Y,Z)=0$ which is if and only if $\sum_{i=1}^n a_i b_j^\ast=0$.
