Let $B$ be a bounded self-adjoint operator on the Hilbert space $(\mathcal{H}, \langle \cdot, \cdot \rangle)$ with $0 \not \in \sigma(B)$ and further let $\rho \in \mathbb{R}$ be strictly positive such that $\rho \not \in \sigma_{ess}(B^{-1})$. Fix some linearly independed vectors $v_1, \dots v_n \in \mathcal{H}$ and define the matrices
$$ A_1 := (\langle v_i, v_j\rangle )_{i,j=1, \dots, n}, $$ $$ A_2 := (\langle Bv_i, v_j\rangle)_{i,j=1, \dots, n}, $$ $$ A_3 := (\langle Bv_i, Bv_j\rangle)_{i,j = 1, \dots, n}. $$
The claim is that $$ A_1 - 2\rho A_2 + \rho^2 A_3 \text{ positive definite} \Leftrightarrow \rho \text{ is not an eigenvalue of } B^{-1}. $$ The "$\Rightarrow$"-statement holds obviously. Unfortunately I'm failing to see why the inverse statement holds. This comes up as an uncommented fact in a paper on a numerical method I'm currently trying to understand. I would appreciate any hint on the problem.
Let $x\in \mathbb R^n$ be given. Set $u:=\sum_{i=1}^n x_iv_i \in H$. Then $$ x^T(A_1-2\rho A_2+\rho A_3)x =\langle u, (I -2\rho B+\rho^2 B^2)u\rangle, $$ since $B$ is self-adjoint and $\rho$ is real. Now, $I -2\rho B+\rho^2 B^2 = (I-\rho B)^2$, and again by self-adjointness $$ x^T(A_1-2\rho A_2+\rho A_3)x =\langle u, (I -2\rho B+\rho^2 B^2)u\rangle =\|(I-\rho B)^2u\|^2 = \rho^2 \|(\rho^{-1}I- B)^2u\|^2. $$ This easily shows, that if $\rho^{-1}$ is not an eigenvalue of $B$ then the matrix is positive definite.
The other direction is false: Consider the diagonal matrix $H=diag(1,\dots, 1, \rho)\in \mathbb R^{n+1,n+1}$. Hence, $\rho$ is an eigenvalue of $B$ with eigenvector $e_{n+1}$. Take the basis $v_i=e_i$ to be the standard basis. This yields $A_1=A_2=A_3=-I_n$, and $A_1-2\rho A_2+\rho A_3=(I-\rho)^2$. If $\rho\ne1$ then this matrix is positive definite but $\rho^{-1}$ is an eigenvalue of $B^{-1}$. The problem is that the chosen basis is orthogonal to the eigenspace of $B$ associated with the eigenvalue $\rho$.