If $\lambda$ is an eigenvalue of a self-adjoint operator, is $\lambda$ in the resolvent set of $\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}$? Let $A$ be a symmetric linear operator on a $\mathbb R$-Hilbert space $H$ and $\lambda\in\mathbb R$. It's easy to see that $$A\left(\mathcal D(A)\cap{\mathcal N(\lambda-A)}^\perp\right)\subseteq{\mathcal N(\lambda-A)}^\perp\tag1$$ and $$A_\lambda:=\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}$$ is a symmetric linear operator on ${\mathcal N(\lambda-A)}^\perp$ such that $\lambda-A_\lambda$ is injective.

Now assume $A$ is bounded. Then $A_\lambda$ is bounded and self-adjoint, $(\lambda-A_\lambda)^\ast$ is injective and hence $\lambda-A_\lambda$ has dense range.
Are we able to conclude $\lambda\in\rho(A_\lambda)$? All what's left to show is boundedness of $(\lambda-A_\lambda)^{-1}$.

Remark: I only need the claim to be true whenever $\lambda$ is an eigenvalue of $A$, but I don't think that we can infer anything more than already known from that.
EDIT: I've observed the following: If $B$ is a nonnegative linear operator on $H$ and $\varepsilon>0$, then $B+\varepsilon$ is injective and $(B+\varepsilon)^{-1}$ is bounded (see, for example, this answer). Thus, if $C$ is a linear operator on $H$ and $\lambda\in\mathbb R$ such that $(\lambda-\varepsilon)-C$ is nonnegtive for some $\varepsilon>0$, then $\lambda-C$ is injecive and $(\lambda-C)^{-1}$ is bounded.
 A: For ease of notation I will denote $\mathcal N(\lambda-A)$ by $N$ and $\mathcal N(\lambda-A)^\perp$ by $N'$. You then have $H=N\oplus N'$ and the operator $A$ respects this decomposition as $A$ is self-adjoint, so you can write $A=(\lambda\Bbb 1\lvert_{N})\oplus B$ where $B:N'\to N'$. The operator $B$ is completely arbitrary, so long as it is self-adjoint and doesn't have an eigenvalue at $\lambda$.
You now want conditions on $A$ so that the operator $B-\lambda$ is invertible. The conditions you list, like for example $\lambda -A$ being non-negative, are not sufficient to guarantee this. As an example consider $\lambda=0$ and $H=L^2([0,1],dx)\oplus \Bbb R$ and let $A(f,z) = (x\cdot f,0)$. This is a non-negative operator and the operator $B$ is the multiplication with $x$ operator on $L^2([0,1],dx)$, which is not invertible.
You can continue to play this game, finding for a condition $A$ a suitable, completely arbitrary operator $B$ (except for the conditions of being self-adjoint and not having an eigenvalue $\lambda$).
So when is $B$ invertible? The relevant criterium is the following: $B$ is invertible iff $\lambda$ is an isolated point of $\sigma(A)$.
This follows directly from
$$\sigma(B)=\overline{\sigma(A)-\{\lambda\}}.$$
So we should think about why that equation is true. First note that in our decomposition $A=(\lambda\Bbb1\lvert_N)\oplus B$ you have $\tilde\lambda-A= (\tilde \lambda- \lambda)\Bbb1\lvert_{N}\oplus (\tilde \lambda -B)$. This is invertible if and only if both operator summands are invertible, so for $\tilde\lambda\neq\lambda$ you have that $\tilde\lambda -A$ is invertible iff $\tilde\lambda -B$ is invertible, in particular $\sigma(B)$ can differ from $\sigma(A)$ by at most the point $\lambda$. However $\sigma(B)$ must be closed, so if $\lambda$ is not an isolated point of $\sigma(A)$ it must be in $\sigma(B)$. On the other hand if $\lambda$ were a isolated point of $\sigma(A)$ and $\lambda\in \sigma(B)$, then $\lambda$ must be isolated in $\sigma(B)$. But isolated points in the spectrum correspond to eigenvalues, so $B$ would have to have an eigenvalue at $\lambda$, which is forbidden, hence if $\lambda$ is isolated it is not in $\sigma(B)$.
