Solution to Review Question by Qiang Gao, updated at Apr 2, 2018
Chapter 1 Finite-Sample Properties of OLS Section 2 The Algebra of Least Squares ...
Prove that
$$
\begin{align}
&
\text{ Both $ \mathbf{P} $ and $ \mathbf{M} $ are symmetric and idempotent, }
\tag{1.2.9}
\ &
\mathbf{P} \mathbf{X} = \mathbf{X} \quad
\text{(hence the term projection matrix),}
\tag{1.2.10}
\ &
\mathbf{M} \mathbf{X} = \mathbf{0} \quad
\text{(hence the term annihilator).}\tag{1.2.11}
\end{align}
$$
(1) $$ \mathbf{P} $$ is symmetric because
$$
\begin{align}
\mathbf{P}^\intercal
& =
[\mathbf{X} (\mathbf{X}^\intercal \mathbf{X})^{-1}
\mathbf{X}^\intercal]^\intercal
&&
\text{(definition (1.2.7))}
\ & =
(\mathbf{X}^\intercal)^\intercal
[(\mathbf{X}^\intercal \mathbf{X})^{-1}]^\intercal
\mathbf{X}^\intercal
&&
((\mathbf{A} \mathbf{B})^\intercal =
\mathbf{B}^\intercal \mathbf{A}^\intercal)
\ & =
( \mathbf{X}^\intercal)^\intercal
[(\mathbf{X}^\intercal \mathbf{X})^\intercal]^{-1}
\mathbf{X}^\intercal
&&
((\mathbf{A}^{-1})^\intercal =
(\mathbf{A}^\intercal)^{-1})
\ & =
\mathbf{X} (\mathbf{X}^\intercal \mathbf{X})^{-1}
\mathbf{X}^\intercal
&&
((\mathbf{A}^\intercal)^\intercal = \mathbf{A})
\ & =
\mathbf{P}.
&&
\text{(definition (1.2.7))}
\end{align}
$$
(2) $$ \mathbf{M} $$ is symmetric because
$$
\begin{align}
\mathbf{M}^\intercal
& =
( \mathbf{I} - \mathbf{P} )^\intercal
&&
\text{(definition (1.2.8))}
\ & =
\mathbf{I}^\intercal - \mathbf{P}^\intercal
&&
((\mathbf{A} - \mathbf{B})^\intercal =
\mathbf{A}^\intercal - \mathbf{B}^\intercal)
\ & =
\mathbf{I} - \mathbf{P}
&&
\text{($\mathbf{I}$ and $\mathbf{P}$ are symmetric)}
\ & =
\mathbf{M}.
&&
\text{(definition (1.2.8))}
\end{align}
$$
(3) $$ \mathbf{P} $$ is idempotent because
$$
\begin{align}
\mathbf{P}^2
& =
(\mathbf{X} (\mathbf{X}^\intercal \mathbf{X})^{-1}
\mathbf{X}^\intercal) \cdot (\mathbf{X}
(\mathbf{X}^\intercal \mathbf{X})^{-1}
\mathbf{X}^\intercal)
&&
\text{(definition (1.2.7))}
\ & =
\mathbf{X} (\mathbf{X}^\intercal \mathbf{X})^{-1}
(\mathbf{X}^\intercal \mathbf{X})
(\mathbf{X}^\intercal \mathbf{X})^{-1}
\mathbf{X}^\intercal
&&
((\mathbf{A} \mathbf{B}) \mathbf{C} =
\mathbf{A} (\mathbf{B} \mathbf{C}))
\ & =
\mathbf{X} (\mathbf{X}^\intercal \mathbf{X})^{-1}
\mathbf{X}^\intercal
&&
(\mathbf{A}^{-1} \mathbf{A} = \mathbf{I})
\ & =
\mathbf{P}.
&&
\text{(definition (1.2.7))}
\end{align}
$$
(4) $$ \mathbf{M} $$ is idempotent because
$$
\begin{align}
\mathbf{M}^2
& =
(\mathbf{I} - \mathbf{P})(\mathbf{I} - \mathbf{P})
&&
\text{(definition (1.2.8))}
\ & =
\mathbf{I} - \mathbf{P} - \mathbf{P} + \mathbf{P}^2
\ & =
\mathbf{I} - \mathbf{P} - \mathbf{P} + \mathbf{P}
&&
\text{($\mathbf{P}$ is idempotent)}
\ & =
\mathbf{I} - \mathbf{P}
\ & =
\mathbf{M}.
&&
\text{(definition (1.2.8))}
\end{align}
$$
(5) $$ \mathbf{P} \mathbf{X} = \mathbf{X} $$ because
$$
\begin{align}
\mathbf{P} \mathbf{X}
& =
\mathbf{X} (\mathbf{X}^\intercal \mathbf{X})^{-1}
\mathbf{X}^\intercal \cdot \mathbf{X}
&&
\text{(definition (1.2.7))}
\ & =
\mathbf{X} (\mathbf{X}^\intercal \mathbf{X})^{-1}
(\mathbf{X}^\intercal \mathbf{X})
\ & =
\mathbf{X}.
\end{align}
$$
(6) $$ \mathbf{M} \mathbf{X} = \mathbf{0} $$ because
$$
\begin{align}
\mathbf{M} \mathbf{X}
& =
( \mathbf{I} - \mathbf{P} ) \mathbf{X}
&&
\text{(definition (1.2.8))}
\ & =
\mathbf{X} - \mathbf{P} \mathbf{X}
\ & =
\mathbf{X} - \mathbf{X}
&&
( \mathbf{P} \mathbf{X} = \mathbf{X} )
\ & = \mathbf{0}.
\end{align}
$$
Copyright ©2018 by Qiang Gao