Let $H$ be a complex, separable Hilbert space, and $T:H \rightarrow H$ a linear, bounded operator. Assume that $$\sigma(T) = \sigma(T^*) = \{ \lambda \in \mathbb{C}: a \leq |\lambda| \leq b \}$$ for $0< a < b$. Now assume that for $a < |\lambda| < b$ we have that $\lambda$ is an eigenvalue for $T^*$ but $T - \lambda I$ is bounded below.
I'm studying a proof that assures that in this case, $(T - \lambda I)^*$ has infinite dimensional kernel. Why? I'm trying to prove that if an operator is bounded below then its adjoint fulfills this property, but I'm stuck trying it.
My first attempt is note that $\ker ( (T - \lambda I)^*) = \overline{ Im(T - \lambda I)}^\perp$ and try to compute the codimension of $Im(T- \lambda I)$, where the fact that $\lambda$ is an eigenvalue may help.
Anyone can help me? Thank you very much.