Sum of projection matrices For the decomposition, where $I$ is the identity matrix and $P_{0}$ and $P_{1}$ are projection matrices, and the decomposition is:
$I = P_{0}+(P_{1}-P_{0})+(I-P_{1})$
The statement is that all the terms on the right hand side are projection matrices.  I struggle to prove that the middle term will be a projection matrix. It is clearly symmetric, but I can't figure out how to prove idempotent. The cross terms are throwing me off.
My question is really, will all the terms be projection matrices for all projection matrices $P_{0}$ and $P_{1}$, or does it depend on $P_{0}$ and $P_{1}$. 
I'm using this in Cochrans theorem to show independnence
 A: If $P$ and $Q$ are orthogonal projectors, then 
$$
(P-Q)^2=P^2+Q^2-PQ-QP=P+Q-PQ-QP=(P-Q)+2Q-PQ-QP.
$$
So $P-Q$ is an orthogonal projector if and only of $2Q=PQ+QP$.
If $P$ and $Q$ commute, this condition is equivalent to $Q=PQ$, that is, $(I-P)Q=0$. Hence, in this case, $P-Q$ is an orthogonal projection if and only if $\mathrm{ker}(P)\perp\mathrm{im}(Q)$.
Commutativity is not enough. Consider, e.g.,
$$
P=\begin{bmatrix}1&0\\0&0\end{bmatrix},
\quad
Q=\begin{bmatrix}0&0\\0&1\end{bmatrix}.
$$
Clearly,
$$
P-Q=\begin{bmatrix}1&0\\0&-1\end{bmatrix}
$$
is not an orthogonal projector.
A: $P_0(P_0(x)) = P_0(x)$,
$(I-P_1)(I-P_1)(x) = (I-P_1)(x-P_1(x)) = I(x-P_1(x)) - P_1(x-P_1(x)) = Ix -IP_1(x) - P_1(x) + P_1(P_1(x))= x - P_1(x) - P_1(x) + P_1(x) = x-P_1(x) = (I-P_1)(x).$
$(P_1-P_0)(P_1-P_0)(x) = (P_1-P_0)(P_1(x) - P_0(x)) = P_1(P_1(x)-P_0(x)) - P_0(P_1(x)-P_0(x)) = P_1(P_1(x)) - P_1(P_0(x)) - P_0(P_1(x)) -P_1(P_0(x)) +P_0(P_0(x))= P_1(x) + P_0(x) = (P_1+P_0)(x)$
if projection matrices $P_0,P_1$ commute.
