Given the infinite dimensional space of all complex polynomials $p(z)$ does the operator $O$ representing multiplication by $x$ have any eigenvectors?
Since the multiplication operator is $O[p(z)]=xp(z),$ doesn't this show that $x$ is an eigenvalue with eigenvector $p(z)$?
On
I assume you mean $x$ is just a number. If this is the case then yes, look at the map $O-\lambda I$. It is defined by $(O-\lambda I)(p)=O(p)-\lambda p=xp-\lambda p=(x-\lambda)p$. And it is obvious its kernel is non-trivial iff $\lambda=x$. So this is the only eigenvalue.
On
Yes, simple scaling operators are the ultimate in degeneracy: every vector is an eigenvalue with the same eigenvalue. That eigenvalue has an algebraic and geometric multiplicity equal to the dimension of the space; in this case, that is $\omega$. We can choose any basis for the space, and the operator will be diagonal in that basis.
No. If the scalar $\lambda$ were to be an eigenvalue of $O$ there would have to be a polynomial $p(z)$ such that $$ O(p(z)) = zp(z) = \lambda p(z) $$ That can't happen because the degree of $zp(z)$ is greater that the degree of $p(z)$.
The question is a little confusing. I read "$x$" as "$z$". If $x$ is just a number then the operator "multiplication by $x$" is just the operator $xI$ and every vector is an eigenvector with eigenvalue $x$.