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Let $Z$ be a $n \times r$ matrix with rank $r$, let $X$ be an $n \times k$ matrix with rank $k$. Consider the projection matrix $P_Z=Z(Z'Z)^{-1}Z'$. In my lectures notes I came across the following decomposition of $P_Z$ into the sum of two other projection matrices:

$$P_Z=P_{Z \hat{\Pi}}+P_{\Pi_{\perp}}$$ where $\hat{\Pi}=(Z'Z)^{-1}Z'X$, a $r \times k$ matrix.

The problem is that I don't recall what $\Pi_{\perp}$ is. I know that it must be of dimension $r \times r-k$. This fact in combination with the perpendicular symbol suggest that it might be a set of column vectors which span the orthogonal complement of $Z \hat{\Pi}$. But when I try to prove the decomposition I think I need that $\hat{\Pi}'Z' Z\Pi_{\perp}=0$, which I dont think can be deduced from the choice I suggested above.

What should be $\Pi_{\perp}?$

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Since $Z\hat\Pi=Z(Z'Z)^{-1}Z'X$, its columns span the projection of the column space of $X$ onto the column space of $Z$. In particular, they span some subspace of the column space of $Z$, although there’s no guarantee that they’re linearly independent. From context, then, $\operatorname{col}(Z) = \operatorname{col}(Z\hat\Pi) \oplus \operatorname{col}(\Pi_\perp)$, i.e., the column space of $\Pi_\perp$ is a complement of $\operatorname{col}(Z\hat\Pi)$ within $\operatorname{col}(Z)$.

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  • $\begingroup$ Would this indeed imply $\hat{\Pi'}Z'Z \Pi_{\perp}=0$? $\endgroup$ Commented Feb 25, 2018 at 9:36
  • $\begingroup$ Because I don't think this is the case and then I don't see how to prove this, except possibly by some very tedious block matrix inverse steps.. $\endgroup$ Commented Feb 25, 2018 at 9:53

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