Independence of linear combination of multivariate distribution $X$ is multivariate normal distributed with $\mu = (2, 2)$ and $\Sigma = (1, 0 ; 0, 1)$ and I have 2 vectors $A = (1, 1)$ and $B = (1, -1)$. How can I than show that the linear combinations $AX$ and $BX$ are independent?
 A: First of all independence follows from a lack of correlation when two random variables are jointly normal.  Now it isn't hard to see that $AX$ and $BX$ are uncorrelated:
\begin{align}
\text{Cov}(AX, BX) &= \text{Cov}(X_1 + X_2, X_1 - X_2) \\
&= \text{Var}(X_1) - \text{Var}(X_2) \\
&= 0
\end{align}
so we just need to show that $AX$ and $BX$ are jointly normal.  To do that it's enough to show that all linear combinations of $AX$ and $BX$ are normally distributed.  Can you finish the proof? 
A: The definition of joint normality of $W$ and $Z$ is that all linear combinations $aW+bZ$ of $W$ and $Z$ are normal random variables. Since $X$ and $Y$ are jointly normal, this shows that $W=X+Y$ and $Z=X-Y$ are normal random variables. Furthermore, $aW+bZ = (a+b)X+(a-b)Y$ is normal because it is a linear combination of jointly normal random variables $X$ and $Y$. We conclude that $X+Y$ and $X-Y$ are jointly normal random variables. Hence they are independent if their covariance is $0$. I leave it to you to use the bilinearity of the covariance function to determine whether they are indeed independent random variables. 
Look, Ma! No explicit formulas for pdfs used anywhere!
