Given the following two quadratic forms: $$a^2=\mathbf{w'Xw}$$ $$b^2=\mathbf{1'X^{-1}1}$$
And the known relations:
$$a^2b^2=1$$ $$\mathbf{X}=\mathbf{\Sigma}-\lambda\mathbf{R}$$
Where $\mathbf{\Sigma}$ and $\mathbf{R}$ are symmetric positive definite matrices.
Is it possible to find a closed expression for the scalar $\lambda$?
Also known:
$$\mathbf{w'1}=1$$
Notice that $\lambda$ is linear in the equation $a^2 = \mathbf w' (\mathbf \Sigma - \lambda \mathbf R) \mathbf w$.
Since $\mathbf R$ is positiv definite and $\mathbf w\ne 0$ , we have $\mathbf w'\mathbf R \mathbf w > 0$ and $$ \lambda = \frac{\mathbf w'\mathbf\Sigma \mathbf w - a^2}{\mathbf w'\mathbf R \mathbf w}.$$