# 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?

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 $$P_1P_2=P_2$$.
• Thanks for the answer! I have several questions for you. What do you mean by "multiply on the left by $I-P_1$"? I do not see how you go from $P_1P_2+P_2P_1=0$ to $(I-P_1)P_2P_1=0$. – Matthew Feb 7 at 0:24
• I mean exactly that. And $(I-P_1)P_1=0$. – Martin Argerami Feb 7 at 0:25
• Additionally, what do you mean by selfadjoint? Are you multiplying both sides of the inequality by $P^{-1}$? – Matthew Feb 7 at 0:25
• I see now. What property is $(I-P_1)P_1=0$? This seems good to know, and it was not in my notes. – Matthew Feb 7 at 0:28
• It's $P_1^2=P_1$. How do you define "projection"? In your context, "selfadjoint" means "symmetric". And $P^{-1}$ makes no sense, the only invertible projection in the identity. – Martin Argerami Feb 7 at 0:30