# Infinite-dimensional analogue of "positive-negative splitting implies non-degeneracy"

(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?

• Can't you just decompose $A=A_++A_-$ into positive and negative definite parts, using the spectral theorem (already in the finite-dimensional case)? Nov 17, 2021 at 16:18
• @Christian There is an off-diagonal part $B$ which I don't know how to handle unless $A_+$ or $A_-$ is assumed to be invertible. Could you give more details? Nov 17, 2021 at 23:55
• There is an orthogonal decomposition $V=V_+\oplus V_-$ such that $V_{\pm}$ are reducing for $A$, and the restrictions are positive/negative definite (use the spectral theorem). Nov 18, 2021 at 15:30

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.