When is the sum and difference of two projection matrices $P_1$ and $P_2$ a projection matrix? Let $P_1$ and $P_2$ be two projection matrices for orthogonal projections onto $S_1 \in \mathcal{R}^m$ and $S_2 \in \mathcal{R}^m$, respectively. When does $P_1+P_2$ and $P_1-P_2$ result in a projection matrix? Prove it.
I am confident that $P_1+P_2$ is a projection matrix iff $P_1P_2=P_2P_1=(0).$
Similarly, I feel like $P_1-P_2$ is a projection matrix iff $P_1P_2=P_2P_1=P_2.$
However, I do not know how to formally prove either of the above statements. How should I formalize the proof?
 A: If $P_1+P_2$ is a projection, then 
$$
P_1+P_2=(P_1+P_2)^2=P_1+P_2+P_1P_2+P_2P_1.
$$
So $P_1P_2+P_2P_1=0$. Multiply on the left by $I-P_1$ to get $(I-P_1)P_2P_1=0$. So $P_2P_1=P_1P_2P_1$, selfadjoint, which then gives $P_1P_2=P_2P_1$. So $2P_1P_2=0$, and $P_1P_2=0$. 
If $P_1-P_2$ is a projection, then 
$$
P_1-P_2=(P_1-P_2)^2=P_1+P_2-P_1P_2-P_2P_1.
$$
So $P_1P_2+P_2P_1=2P_2$. Multiply by $I-P_2$ on the right, to get $P_2P_1(I-P_2)=0$. As above, we conclude that $P_2P_1=P_1P_2$. Then 
$$
2P_2=P_1P_2+P_2P_1=2P_1P_2,
$$
and $P_1P_2=P_2$. 
A: A similar answer as Martin Argerami’s, but without using selfadjointness:
If $P_1\pm P_2$ is a projection, then
$$
P_1\pm P_2=(P_1\pm P_2)^2=P_1+P_2\pm P_1P_2\pm P_2P_1\;.
$$
and thus
$$
P_1P_2+P_2P_1=P_2\mp P_2\;.
$$
Multiply by $P_2$ from the left to obtain
$$
P_2P_1P_2+P_2P_1=P_2\mp P_2\;.
$$
and from the right to obtain
$$
P_1P_2+P_2P_1P_2=P_2\mp P_2\;.
$$
Subtracting the two equations yields $P_1P_2=P_2P_1$, and thus
$$
P_1P_2=P_2P_1=\frac12\left(P_2\mp P_2\right)\;.
$$
Note that the last step does not work over a field of characteristic $2$, and indeed over $\mathbb F_2$ we have the counterexample
$$
P_1=P_2=\pmatrix{1&0\\0&0}\;,
$$
where $P_1+P_2=0$, a projection matrix, despite $P_1P_2=P_1\ne0$.
