Let $H=(H, (\cdot, \cdot ))$ be a Hilbert space over $\mathbb{R}$ and $A : D(A) \subset H \longrightarrow H$, with $\overline{D(A)}=H$, a linear self-adjoint operator. Suppose $ A $ has only a negative eigenvalue $\lambda<0$, with eigenvector associated $v \in D(A)$, then we have $A(v)=\lambda v$.
Question. For all $w \in D(A)$ such that $(w,v)=0$ we have $$(A(w),w) \geq 0?$$
What I got to think is: If $w \in D(A)$ is such that $(w,v)=0$ then $$0=(w,\lambda v)=(w, A(v))= (A(w),v) $$ but I could not conclude anything.
The answer is no for a general linear operator, yes for a compact operator. The reason is that $A$ could have non-point spectrum on the negative real axis (this is not possible if $A$ is compact).
For a concrete example, consider $H = L^2(-1,1), A(u)(x) = x(u(x) - \bar u) - \bar u$, where $\bar u = \frac{1}{2} \int_{-1}^1 u(x) dx$ is the average. Then $A$ is bounded and self-adjoint and has the eigenvalue $\lambda = -1$ with eigenfunction $v(x) = 1$ and no other eigenvalues. Now set $$ w(x) = \begin{cases} 1 \quad (x > -1/2) \\ -3 \quad (x \le - 1/2) \end{cases} $$
Then $\langle w, v \rangle = 0 = \bar w$ and $Aw(x) = \begin{cases} x \quad (x > -1/2) \\ -3x \quad (x \le - 1/2) \end{cases}$. Therefore $$ \langle Aw,w \rangle = \int_{-1}^{-1/2}9x dx + \int_{-1/2}^1 x dx \, .= -3 $$