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Yes IV's don't change when you power them up or even evolve them.

A 100% IV Pokemon will stay 100% no matter what you do to them.

Having a higher IV will mean that the CP cap is higher.

Let me expand on the setting. Let $\text{Hol}(T)$ be the set of all the functions holomorphic in a neighborhood of $\sigma(T)$. The Riesz functional calculus now says that indeed if I have the polynomials $p$, that $p(T)$ is as expected - of course, for polynomials this is trivial.

Additionally, we have the Banach algebra of bounded operators on $X$, and: Spectral mapping theorem: If $T$ is an element of a Banach algebra $\mathcal A$ and $f$ is holomorphic in a neighborhood of $\sigma(T)$, then $$\sigma(f(T)) = f(\sigma(T)).$$

If we thus select such a polynomial $p$ such that $p(T) = 0$, we get $p(\sigma(T)) = 0$.

It is enough to show the contrapositive, namely, that if $P(\lambda) \neq 0$, then $T - \lambda \operatorname{id}_X$ is invertible. Now, since $P(\lambda) \neq 0$, $x-\lambda$ does not divide the polynomial $P(x)$, so that we may write $P(x) = Q(x)(x-\lambda) + R$ for some polynomial $Q(x)$ and a non-zero remainder $R \in \mathbb{C}$ — since $x-\lambda$ is of degree one, the remainder must be of strictly lower degree, and thus constant. But then, this means that $0 = P(T) = Q(T)(T-\lambda \operatorname{id}_X) + R\operatorname{id}_X$, so that since $T-\lambda \operatorname{id}_X$ and $Q(T)$ commute, we find that $T-\lambda \operatorname{id}_X$ is invertible with bounded inverse $(T-\lambda \operatorname{id}_X)^{-1} = -\tfrac{1}{R}Q(T)$, as required.