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Let $B$ a bounded self-adjoint operator on a Hilbert space $H=\{f: L^2(\mathbb{R}^n)\rightarrow \mathbb{R}\}$ with an associated inner product $(\cdot,\cdot).$ Take $V=span\{f_1, f_2, \ldots, f_n\}$ a finite dimensional subspace of $H$. Prove that the dimension of the maximal subspace of $V$ on which B is positive definite (resp. negative definite) equals the number of positive (resp. negative) eigenvalues of the matrix $M\in \mathbb{R}^{nxn}$ with elements $m_{ij}=(B f_i, f_j)$ for all $1\leq i,j\leq n.$

I realize that this matrix comes from the fact that if $w=c_1f_1+\cdots+c_nf_n$ with all $c_i \in \mathbb{R}$, then $(Bw,w)=(c_1 \ldots c_n)M(c_1 \ldots c_n)^T.$ Also, $M$ has only real eigenvalues since $B$ is self-adjoint and that $H$ could be taken a more general Hilbert space. But I do not know how to proceed from here. Thank you for any help.

Frank Zermelo
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    The matrix $M$ corresponds to the operator $PMP,$ where $P$ is the orthogonal projection onto $V.$ This should help. In my opinion $L^2$ can be replaced by any Hilbert space. – Ryszard Szwarc Nov 11 '24 at 18:15
  • Since B is already restricted on V, I don't see how the projection on V would help. If you have a little more details, I would be grateful. – Frank Zermelo Nov 11 '24 at 20:35

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