# Determine if orthogonal projection and full rank

Let $$X=(X_1,X_2)$$ be a $$n \times p$$ matrix of rank $$p$$, where $$X_1$$ is $$n\times p_1$$ and $$X_2$$ is $$n\times p_2$$. Let $$P_1$$ be orthogonal projection onto $$C(X_1)$$ and $$P_2$$ be orthogonal projection onto $$C(X_2)$$. Now define $$U_2=(I-P_1)X_2$$, and let $$P_2^*=(U_2)(U_2^T(U_2))(^-)(U_2^T)$$. Finally, let $$P=P_1+P_2^*$$.

Are $$P$$ and $$P_2^*$$ orthogonal projections?

Is $$U_2$$ full rank?

Define $$U_1=(I-P_2)X_1$$ and $$P_1^*=(U_1)(U_1^T(U_1))(^-)(U_1^T)$$. Let $$Q=P_2+P_1^*$$. Does $$Q=P$$?

• If $P=P_1+P_2^*$, and both $P_1$ and $P_2^*$ are projections, there is no way $P$ and $P_2^*$ are going to be orthogonal, and $P$ will typically not even be a projection. – Thomas Lumley Oct 5 '20 at 6:06
• @Thomas In this setting, $P_1$ and $P_2^{*}$ will be orthogonal projections by construction. – whuber Oct 5 '20 at 14:54
• Yes, but he asked if $P$ and $P_2^*$ are orthogonal, where $P=P_1+P_2^*$, and those aren't orthogonal. It's true that $P$ will be a projection, though. – Thomas Lumley Oct 5 '20 at 21:10