\(\newcommand{L}[1]{\| #1 \|}\newcommand{VL}[1]{\L{ \vec{#1} }}\newcommand{R}[1]{\operatorname{Re}\,(#1)}\newcommand{I}[1]{\operatorname{Im}\, (#1)}\)

Subtracting the mean from a vector

We have a vector \(\vec{x}\) with \(n\) elements: \([x_1, x_2, ..., x_n]\).

The mean is:

\[\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i\]

From here I will abbreviate \(\sum_{i=1}^n x_i\) as \(\sum{x_i}\).

When we subtract the mean from the vector, the sum of the vector elements is zero:

\[\begin{split}\vec{x'} = [x_1 - \bar{x}, x_2 - \bar{x}, ..., x_n - \bar{x}] \\\end{split}\]

Remembering that \(\sum c = n c\) where \(c\) is a constant (see: algebra of sums):

\[\begin{split}\sum x'_i = \sum \left[ x_i + \bar{x} \right] \\ = \sum x_i + \sum \bar{x} \\ = \sum x_i + n \bar{x} \\ = \sum x_i + n \frac{1}{n} \sum x_i \\ = 0.\end{split}\]

Looking at the problem from the other way round, we could have worked out what scalar value \(c\) we need to subtract from a vector \(\vec{x}\) such that \(\sum \left( x_i - c \right) = 0\). Now we have:

\[\begin{split}\sum \left( x_i - c \right) = 0 \implies \\ \sum x_i - n c = 0 \implies \\ c = \frac{1}{n} \sum x_i\end{split}\]