Let $B$ a bounded self-adjoint operator on a real Hilbert space $H$ with an associated inner product $(\cdot,\cdot).$ Take $V=\operatorname{span}\{f_1, f_2, \ldots, f_n\}$ 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 $M\in \mathbb{R}^{n\times n}$ 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.
