(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?

  • $\begingroup$ Can't you just decompose $A=A_++A_-$ into positive and negative definite parts, using the spectral theorem (already in the finite-dimensional case)? $\endgroup$ Nov 17, 2021 at 16:18
  • $\begingroup$ @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? $\endgroup$
    – Xin Nie
    Nov 17, 2021 at 23:55
  • $\begingroup$ 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). $\endgroup$ Nov 18, 2021 at 15:30

1 Answer 1


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.


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