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In the Generalized Ensemble Method section of this paper, the authors gave an optimization problem, which is

\begin{equation} \min \sum_{i=1}^N\sum_{j=1}^N\alpha_i\alpha_jC_{ij} \\ \text{s.t.}\quad \sum_{i=1}^N \alpha_i=1 \\ \end{equation} By using Lagrange multiplier and setting derivatives to zero, we get \begin{equation} \sum_{j=1}^N\alpha_jC_{ij}=\lambda,~~~i=1,2,\dots,N \\ \sum_{i=1}^N \alpha_i=1 \end{equation}

Now my question is how to manipulate the above equations and get the solution $\alpha_{i}=\frac{\sum_{j} C_{i j}^{-1}}{\sum_{k} \sum_{j} C_{k j}^{-1}}$ ?

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Calling $\mathcal{1}= (1,\cdots, 1)$ the Lagrangian can be stated as

$$ L = \alpha^{\dagger}\cdot C\cdot \alpha + \lambda (\alpha\cdot \mathcal{1}-1) $$

and the stationary points are the solution to

$$ \nabla L = 2 C\cdot \alpha + \mathcal{1}\lambda = 0 $$

now assuming $C$ invertible

$$ \alpha^* = -\frac 12 C^{-1}\cdot \mathcal{1}\lambda $$

but

$$ (\alpha^*)^{\dagger}\cdot \mathcal{1} = 1 = -\frac 12\lambda \mathcal{1}^{\dagger}\cdot C^{-1}\cdot \mathcal{1} $$

then

$$ \lambda^* = -\left(\frac 12\mathcal{1}^{\dagger}\cdot C^{-1}\cdot \mathcal{1}\right)^{-1} $$

and finally

$$ \alpha^* = \frac{C^{-1}\cdot \mathcal{1}}{\mathcal{1}^{\dagger}\cdot C^{-1}\cdot \mathcal{1}} $$

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