Literally, let $H : \mathcal{H} \to \mathcal{H}$ be an operator of Hilbert space, or finite dimensional inner product space. Assume that $H$ has an eigenvector $\lambda$ with unit eigenvector $v$. Then $(H-\lambda)(\{v \}^{\perp}) \subseteq \{ v\}^{\perp}$ ( the orthogonal complement of $v$ ) and the restriction of $H-\lambda$ to $\{ v\}^{\perp}$ is invertible on that space? If this is not true in general, then when is true? How about the case that $H$ is self adjoint ( or normal )? I have been struggled with this for hours and not prove until now. Is there a counter example?
P.s. My question originates from following argument in Folland, Quantum field theory, a tourist guide for mathematician, p.127 :
Why the underlined statement is true?
Can anyone help?
The statement is true if $H$ is a normal operator on a finite-dimensional Hilbert space and $\lambda$ is a simple eigenvalue, i.e. the null space of $H - \lambda$ is one-dimensional. Counterexamples when $H$ is not normal or $\lambda$ is not simple are easy to find on $2 \times 2$ matrices.
If the Hilbert space is infinite-dimensional, it can happen that the inverse of the restriction of $H-\lambda$ to $\{v\}^\perp$ is unbounded, so in that sense this restriction is not "invertible".