We can center an n-dimensional vector $x$ by calculating $c(x) = x - \mu(x)$ (where $\mu(x)$ is the expectation of the vector x). There is a claim (that I don't understand) that "This can equivalently be seen as a (non-orthogonal) linear projection onto the hyperplane through the origin orthogonal to the vector $\mathbf{1}=(1,1,\ldots,1)$. "
This is not obvious to me at all - can someone help provide a proof (and preferably some intuition) to see why this is the case? Thanks!
That hyperplane has equation $\sum_{i=1}^n y_i = 0$. Now take $y_i = x_i - \mu$ and notice that $$\sum_{i=1}^n y_i = \sum_{i=1}^n (x_i - \mu) = \sum_{i=1}^n x_i - \sum_{i=1}^n \mu = n \mu - n \mu = 0,$$ so the vector $(x_i-\mu)_{i=1}^n$ is indeed in the hyperplane.