Given $\textbf{P} = \textbf{X}(\textbf{X}^{\prime}\textbf{X})^{-1}\textbf{X}^{\prime}$, prove that $\mathcal{C}(\textbf{P}) = \mathcal{C}(\textbf{X})$

Question

If $\textbf{X}\in\textbf{R}^{n\times p}$ has full rank ($n\geq p$), so that $\textbf{P} = \textbf{X}(\textbf{X}^{\prime}\textbf{X})^{-1}\textbf{X}^{\prime}$, prove that $\mathcal{C}(\textbf{P}) = \mathcal{C}(\textbf{X})$, where $\mathcal{C}(\textbf{X})$ indicates the column space of $\textbf{X}$.

I do not need the answer necessarily. Any hints on the problem would be helpful. Thanks in advance.

Answer 1

$\newcommand{\C}{\mathcal{C}}\newcommand{\P}{\mathbf{P}}\newcommand{\X}{\mathbf{X}}$Hint: $\C(\P)\subseteq \C(\X)$ should be clear (look at the form of $\P$). To show that $\C(\X)\subseteq \C(\P)$, note that $\P\X = \X$.