by Qiang Gao, updated at May 17, 2017
...
Let $$ \mathbf{1} $$ be the
(a) $$ \mathbf{M}_1 $$ is symmetric and idempotent.
(b) $$ \mathbf{M}_1 \mathbf{1} = \mathbf{0} $$.
(c) $$ \mathbf{M}_1 \mathbf{y} = \mathbf{y} - \bar{y} \cdot \mathbf{1} $$ where
$$ \mathbf{M}_1 \mathbf{y} $$ is the vector of deviations from the mean.
(a) The symmetry of $$ \mathbf{M}_1 $$ is verified as
The idempotency of $$ \mathbf{M}_1 $$ is verified as
$$ \begin{align} \mathbf{M}_1 \mathbf{M}_1 & = ( \mathbf{I}_n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' )( \mathbf{I}_n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' ) \ & = \mathbf{I}_n
- \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
- \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
- \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \ & = \mathbf{I}_n
- \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
- \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
- \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \ & = \mathbf{I}_n
- \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \ & = \mathbf{M}_1. \end{align} $$
(b)
(c)
$$ \begin{align} \mathbf{M}_{1} \mathbf{y} & = ( \mathbf{I}n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' ) \mathbf{y} \ & = \mathbf{y} - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \mathbf{y} \ & = \mathbf{y} - \mathbf{1} \cdot \frac{1}{n} \cdot \sum{i=1}^{n} y_i \ & = \mathbf{y} - \bar{y} \cdot \mathbf{1}. \end{align} $$
Copyright ©2017 by Qiang Gao