(This question is related to [Splitting a space into positive and negative parts](https://mathoverflow.net/questions/7709/splitting-a-space-into-positive-and-negative-parts) but different.)


Given a finite-dimensional vector space $V$ over $\mathbb{R}$, what I call a "positive-negative splitting" for a symmetric bilinear form $\langle\cdot,\cdot\rangle$ on $V$ is a splitting $V=V\_+\oplus V\_-$ (not necessarily $\langle\cdot,\cdot\rangle$-orthogonal) such that the restrictions of $\langle\cdot,\cdot\rangle$ to $V\_+$ and $V\_-$ are positive definite and negative definite, respectively.


If there exists such a splitting, then $\langle\cdot,\cdot\rangle$ is written in matrix form as
$$
\langle x,y\rangle=
x^\mathsf{T}
\begin{pmatrix}
A\_+&B\\
B^\mathsf{T}&A\_-
\end{pmatrix}
y,
$$
where $A\_+$ and $A\_-$ are positive and negative definite symmetric matrices, respectively. We can calculate the determinant of the above block matrix by Gauss elimination and get
$$
\det
\begin{pmatrix}
A\_+&B\\
B^\mathsf{T}&A\_-
\end{pmatrix}=\det(A\_+)\det(A\_-\!-B^\mathsf{T}A\_+^{-1}B)\neq0
$$
(note that $A\_-\!-B^\mathsf{T}A\_+^{-1}B$ is negative definite). So $\langle\cdot,\cdot\rangle$ is non-degenerate in this case.


My question is:


**For a Hilbert space $(\mathcal{H},(\cdot|\cdot))$ and a bilinear form $\langle\cdot,\cdot\rangle:=(A\cdot|\cdot)$ on $\mathcal{H}$ given by a self-adjoint operator $A:\mathcal{H}\to\mathcal{H}$, does the existence of a positive-negative splitting still imply that $\langle\cdot,\cdot\rangle$ is non-degenerate?**



Infinite-dimensional analogue of "positive-negative splitting implies non-degeneracy"
A simple proof:


Assume $\mathcal{H}=\mathcal{H}\_+\oplus\mathcal{H}\_-$, with $( A u| u)>0$ (resp. $<0$) for all nonzero $u\in\mathcal{H}\_+$ (resp. $u\in\mathcal{H}\_-$). We want to show $\ker A=0$. To this end, assume $Au=0$ and write $u=u\_++u\_-$ with $u\_\pm\in\mathcal{H}\_\pm$. If either $u\_+$ or $u\_-$ is zero, the assumption easily implies $u=0$. Otherwise, we have
$$
0=(Au|u\_+)=(Au\_+|u\_+)+(Au\_-|u\_+)>(Au\_-|u\_+),
$$
$$
0=(Au|u\_-)=(Au\_+|u\_-)+(Au\_-|u\_-)<(Au\_+|u\_-).
$$
In view of the self-adjointness, we get a contradiction.


