<s>[INST] Help with Linear Transformation of a multivariate normal Given X ~ $N_2$ (μ, Σ)$
Find the Distribution of
$$
        \begin{pmatrix}
        X+Y \\
        X-Y
        \end{pmatrix}
$$
Show independence if $Var(X) = Var(Y)$
Attempt:
Given proper of Multitvariate Normal Transformations $N_m$ (Aμ, $AΣA^t$ )
Using $$     A =   
        \begin{pmatrix}  
        1&1 \\       
        1 &-1
        \end{pmatrix}*  \begin{pmatrix}  
        X \\       
        Y
        \end{pmatrix} =
        \begin{pmatrix}
        X+Y \\
        X-Y
        \end{pmatrix}\begin{pmatrix}
         1&1 \\       
        1 &-1
        \end{pmatrix} =
$$
The general Variance-covariance matrix of a bivariate normal,multiplied by $AΣA^t$
$$     A =   
        \begin{pmatrix}  
        1&1 \\       
        1 &-1
        \end{pmatrix}*  \begin{pmatrix}  
        δ_x^2&δ_{xy} \\       
        δ_{xy} &δ_y^2
        \end{pmatrix} *
        \begin{pmatrix}
         1&1 \\       
        1 &-1
        \end{pmatrix} =
\begin{pmatrix}
         4δ&0 \\       
        0 &0
        \end{pmatrix} $$
Given  $δ_{xy} = δ_x^2 = δ_y^2$
My Questions: 
i. Do the zeros in the final Variance-Covariance matrix sufficiently show independence?
ii. Does $δ_y^2 = 0$ effect the pdf, ie making it degenerate?
Where the bivariate pdf is
$\frac{1}{2\piδ_yδ_x(1-\rho^2)} e^{\frac{q}{2(1-\rho^2)}}$
$q = [(\frac{x-μ_x}{δ_x})^2 -2\rho(\frac{y-μ_y}{δ_x})(\frac{y-μ_y}{δ_y}) + (\frac{y-μ_y}{δ_y})^2]$
By $AΣA^t$  $\rho = 0$ and $δ_y^2 = 0 $
$q = [(\frac{x-μ_x}{δ_x})^2 + (\frac{y-μ_y}{0})^2]$  which is problematic.
I'm not sure where I am going wrong here...
 [/INST] *

*Yes, for a Gaussian random vector $(X,Y)$, it is know that $X$ and $Y$ are independent if and only if the covariance $\text{cov}(X,Y)$ equals $0$. Since the entries on the off-diagonal are exactly the covariance, this means that $X$ and $Y$ are indeed independent.

*Well, if $\delta_y^2=0$ (which is equivalent to $\text{var}(Y)=0$), then $Y$ is degenerate Gaussian, i.e. $Y=\mathbb{E}Y$ almost surely.

</s>