(This question is related to 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?**