In constructing a Brownian motion in the book with the same name by Mörters and Peres, they come to a point on page 10 where they want to show that $\{B(d)-B(d-2^{-n})\}_{d\in \mathcal{D}_n}$ are independent. Here $\mathcal{D}_n = \{i/2^n, i = 0,\dotsc, 2^n\}$ and $B$ is constructed iteratively so that on step $n$ $B(d)-B(d-2^{-n})$ has normal distribution with mean $0$ and variance $2^{-n}$. They say:
that it suffices to show that they are pairwise independent, as the vector of these increments is Gaussian.
I see how independence follows if $X = (B(d_1)-B(d_1-2^{-n}), B(d_2)-B(d_2-2^{-n}),\dotsc, B(d_{\mathcal{|D_n|}}-B(d_{|\mathcal{D}_n|}-2^{-n})))$ is shown to be a Gaussian vector. But for this to be the case we need to find a standard Gaussian vector $Y$, a matrix $A$ and a constant vector $b$ such that $X = AY+b$. I'm wondering how this can be done.
$A$ is a diagonal matrix with diagonal entries ${2^{-1/2},...,2^{-n/2}}$.
The variance of $X=AY$ is $YY'$, a diagonal matrix with diagonal entries ${2^{-1},...,2^{-n}}$.