When is the sum and difference of two projection matrices $P_1$ and $P_2$ a projection matrix?

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 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 $P_1P_2=P_1\ne0$.