Show that if $X$ $=$ [$X_1$ $X_2]$ and $X_1'X_2 = 0$, then $P = P_1 + P_2$, where $P$ is defined as $X(X'X)^{-1}X'$, the projection matrix.
Don't quite know where to start. I tried evaluating it by brute force, but there must be an easier way, surely? Don't quite want to deal with taking the inverse of a partitioned matrix. Any help appreciated.
Just simply put $X=[X_1 X_2]$ into $X(X^TX)^{-1}X^T$. $P=[X_1 \ \ X_2]\begin{bmatrix} (X_1^TX_1)^{-1} & 0\\ 0 & (X_2^TX_2)^{-1} \end{bmatrix}\begin{bmatrix} X_1^T\\ X_2^T \end{bmatrix}=X_1(X_1^TX_1)^{-1}X_1^T + X_2(X_2^TX_2)^{-1}X_2^T = P_1 + P_2$. Because $X_1^TX_2=0$, so the inverse is for an block-diagonal matrix, which is easy to compute analytically.