In a $C^*$ algebra if we consider a normal element, say $x$, such that spectrum of $x$ is $\left\lbrace-1,1 \right\rbrace$,then can we find two non-trivial projections $p$ and $q$ such that $pq=0$?
I'm trying to figure out the answer. My approach is - As $x$ is normal and spectrum contains $\left\lbrace -1,1 \right\rbrace$, so x is self adjoint, so we can write is as combination of two unitary element. I don't know if the approach is wrong or right and how to proceed further.
Since the spectrum is discrete, you can take any two continuous functions $f,g$ with $f(-1)=1$, $f(1)=0$, $g(-1)=0$, $g(1)=1$. Then take $$ p=f(x),\ \ \ q=g(x). $$ By spectral mapping you have that $\sigma(p)=\{0,1\}=\sigma(q)$, so both are projections. And as $f(t)g(t)=0$ on $\sigma(x)$, you get $pq=f(x)g(x)=fg(x)=0$.
Finally, since $g(t)-f(t)=t$ on $\sigma(x)$, you get $$ x=q-p. $$
Note that $x$ is actually a self-adjoint unitary. Now let $$p=\frac12(1+x),\qquad q=1-p.$$ Then $p$ is a projection, as is $q$, and $pq=p(1-p)=0$.
