# Spectral properties of operators mapped to zero by some polynomial

Let $$T$$ be a bounded operator on a Banach space $$X$$ and suppose that there is a non-constant polynomial $$p$$ such that $$p(T) = 0$$. It seems to be well known that the spectrum of such an operator coincides with the point spectrum and consists of finite order poles of the resolvent, only. But I can not find any citable reference for this fact.

However, it is not hard to see for me that the spectrum coincides with the point spectrum and that is contained in the set of zeroes of $$p$$ and hence isolated. But I have no clue, why each of this spectral point is a pole of finite order of the resolvent of $$T$$.

Take the annihilating polynomial and shift its argument by $$\lambda I$$. That is, define the polynomials $$q$$ and $$q_0$$ by the identity $$(T-\lambda I) q(\lambda,T-\lambda I) - q_0(\lambda) I = p((T-\lambda I) + \lambda I).$$ Then your hypothesis $$p(T)=0$$ implies the following formula for the resolvent: $$(T-\lambda I)^{-1} = \frac{q(\lambda, T-\lambda I)}{q_0(\lambda)}.$$ It is now obvious that the poles of the resolvent only come from the vanishing of the scalar polynomial denominator $$q_0(\lambda)$$, which I think should be sufficient for your purposes.
• Why does $q$ have two arguments? Is that a typo? – Yaddle Mar 6 at 19:32
• @Yaddle, $q$ separately polynomially depends on $\lambda$ and $T-\lambda I$. More generally $p(x+y) = x q(y,x) - q_0(y)$, the way I defined things. – Igor Khavkine Mar 6 at 19:36