1) Let's say we have the linear model $y=Xβ$. Suppose the model has full rank $X$ and projection matrix $P_X$, show that $P_XX = X$ and that $C(P_X) = C(X)$.
2) Suppose that $P_1$ and $P_2$ are projections for the model $Q_1$ nested within model $Q_2$. Show that $P_1$, $P_2 - P_1$, and $I-P_1$ are all projection matrices and are all mutually orthogonal.
I've been having some real difficulties with these 2 problems. I think the first one might be simple (I have a guess for the column space equivalency) but I don't even know where to start for the second one-the mutual orthogonality is what gets me. I'd appreciate any help I can get.
I'm not sure that the following answer will answer all you questions, but I guess it can help.
1) For a model of a kind $y=X\beta$ and full rank (i.e., there are no linear dependencies between the columns of $X$), you know that the columns of $X$ construct a basis for $C(X)$.
So you can construct the projection matrix onto $C(X)$ using $X$ itself. Such a matrix will be of the following form $P_X=X(X'X)^{-1}X'$. Note that the full rank is essential for $(X'X)^{-1}$ to exist. So, you can easily verify that $P_X$ is indeed an orthogonal projection by checking that $P_X^2=P_X$ and $P_X^T=P_X$. What is the geometrical/intuitive sense of an orthogonal projected vector $y$ onto $C(X)$? The projected vector $\hat{y}$ is the closest vector in $C(X)$ to $y$. So, what will be the closest vector to some $x\in C(X)$ in $C(X)$? The $x$ itself. So, Mathematical is trivial to show that $P_XX=P_X$ because $X(X'X)^{-1}X'X=XI=X$, and intuitively $P_XX$ just projects every column in $X$ onto $C(X)$, hence it returns the $X$ itself. The rank of $P_X$ equals the dimension of $C(X)$ i.e., $p$. And $P_X$ is constructed by the columns of $X$. Hence, both the columns of $X$ and $P_X$ span the same subspace.
2) Moreover, by the very construction of $P_X$ you require orthogonality between non zero vector $y\in \mathbb{R}^n$ and $y-\hat{y}$. Or formally, rearrange $y=\hat{y}=P_Xy$ such that $$P_X^T(y-\hat{y})=P_X^T(y-P_Xy)=P_X^T(I-P_X)y= 0. $$
You can easily check that $I-P_X=I-X(X'X)^{-1}X'$ is indeed symmetric and idempotent matrix, hence it is an orthogonal projection and as you can see, it projects onto the orthogonal complement of $C(X)$, i.e., $C(X)^\perp$. Thus for every $x\in C(X)$, $(I-P_X)x=0$. For $P_2-P_1$ is the same principle, just show that $(P_2-P_1)^T=(P_2-P_1)^T$ and $(P_2-P_1)^2=(P_2-P_1)$.