Sum and difference of projectors

Question

Assume $A$ and $B$ are projectors in $\mathbb{R}^n$. Prove that if $A + B = 0$ then $A - B$ is a projector.

What I've done: we need to prove that $$(A - B)(A - B) = A - B$$ $$A - AB - BA + B = -AB - BA$$ because $A + B = 0$. Now multiple $A + B = 0$ by $A$ and $B$ and sum, it will be $$A + BA + AB + B = 0$$ hence $AB + BA = 0$ therefore in order to prove that $A - B$ is a projector I should prove that $A - B = 0$ and here I got stuck.

Could you please help me how to proceed? Thanks a lot in advance!

Answer 1

Your argument kind of starts with the conclusion.

What you know that is that $A+B=0$. In particular, since $A,B$ are projections, $$\tag1 0=(A+B)^2=A^2+B^2+AB+BA=A+B+AB+BA=AB+BA. $$ Now, since $(I-A)A=0$, $$ 0=(I-A)(A+B)=(I-A)B=B-AB. $$ So $AB=B$. Similarly, $$ 0=(A+B)(I-A)=B(I-A)=B-BA, $$ so $B=BA$. Then, from $(1)$, $$ 0=AB+BA=B+B=2B. $$ Then $B=0$. Now $0=A+B=A+0=A$, so $A=0$.

Answer 2

If $A+B=0$, then $B=-A$ which cannot be a projector unless $B=A=0$. As a principle $p$ being a projector, if $-p$ is a projector then $p=0$. Indeed, looking at the spectra, if $B$ is a projector then $spec(B)\subset \{0,1\}$ and $spec(−B)=spec(A)\subset \{0,−1\}\cap\{0,1\}$. Hence spec(B)={0} but $B$ being a projector this implies $B=0$.