Does the operator $Af(x)= x^2f(x)$ on $H = L^2([-1,4])$ have a cyclic vector?

Question

Let $H = L^2([-1,4])$ and let $A:H\to H$ be defined by $$ Af(x)= x^2f(x). $$ Does this operator have a cyclic vector? I know that having a cyclic vector relates to having eigenvalues of algebraic multiplicity $1$. I'm not familiar with algebraic and geometric multiplicity in infinite dimensional spaces. It seems $A$ does not have a cyclic vector because elements of $\sigma(A) = [0,16]$ in the subinterval $[0,1]$ have algebraic multiplicity 2. Is this correct?