M0M_0: Deviation from Mean Matrix

Proposition

Let x∈Rn×1\pmb x \in \mathbb{R}^{n\times 1} then we can construct a symmetric and idempotent matrix that provides deviations from the mean for a vector of inputs.

This matrix takes is the form:

M0=I−ιι′n\begin{align} M_0 = I - \frac{\pmb \iota \pmb \iota'}{n} \end{align}

where ι\pmb \iota is a vectors of 11 in RN\mathbb{R}^N.

Proof

Symmetric

M0′=(I−ιι′n)′=I′−(ιι′)′n=I−ι′′ι′n=I−ιι′n=M0\begin{align} M_0' &= \left(I - \frac{\pmb \iota \pmb \iota'}{n}\right)' \\ &= I' - \frac{(\pmb \iota \pmb \iota')'}{n} \\ &= I -\frac{\pmb \iota'' \pmb \iota'}{n} \\ &= I - \frac{\pmb \iota \pmb \iota'}{n} \\ &= M_0 \end{align}

Idempotence

M0M0=(I−ιι′n)(I−ιι′n)=I(I−ιι′n)−ιι′n(I−ιι′n)=(I−ιι′n)−ιι′n+ιι′n=I−ιι′n\begin{align} M_0M_0 &= \left(I - \frac{\pmb \iota \pmb \iota'}{n}\right)\left(I - \frac{\pmb \iota \pmb \iota'}{n}\right) \\ &= I\left(I - \frac{\pmb \iota \pmb \iota'}{n}\right) - \frac{\pmb \iota \pmb \iota'}{n}\left(I - \frac{\pmb \iota \pmb \iota'}{n}\right)\\ &= \left(I - \frac{\pmb \iota \pmb \iota'}{n}\right)- \frac{\pmb \iota \pmb \iota'}{n} + \frac{\pmb \iota \pmb \iota'}{n} \\ &= I - \frac{\pmb \iota \pmb \iota'}{n} \end{align}

Deviations From the Mean

M0x=(I−ιι′n)x=x−[ι′x/nι′x/n⋮ι′x/n]=x−xˉ=x¨\begin{align} M_0 \pmb x &= \left(I - \frac{\pmb \iota \pmb \iota'}{n}\right)\pmb x \\[10pt] &= \pmb x - \begin{bmatrix} \pmb \iota' \pmb x/n \\ \pmb \iota' \pmb x/n \\ \vdots \\ \pmb \iota' \pmb x/n \end{bmatrix} \\ &= \pmb x - \bar {\pmb x} \\ &= \ddot{\pmb x} \end{align}