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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?

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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 $P_1P_2=P_2$.

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  • $\begingroup$ 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 $(I-P_1)P_2P_1=0$. $\endgroup$ – Matthew Feb 7 at 0:24
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    $\begingroup$ I mean exactly that. And $(I-P_1)P_1=0$. $\endgroup$ – Martin Argerami Feb 7 at 0:25
  • $\begingroup$ Additionally, what do you mean by selfadjoint? Are you multiplying both sides of the inequality by $P^{-1}$? $\endgroup$ – Matthew Feb 7 at 0:25
  • $\begingroup$ I see now. What property is $(I-P_1)P_1=0$? This seems good to know, and it was not in my notes. $\endgroup$ – Matthew Feb 7 at 0:28
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    $\begingroup$ 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. $\endgroup$ – Martin Argerami Feb 7 at 0:30

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