# Minimize a quadratic form on the 2d probability simplex For a symmetric $$2 \times 2$$ matrix $$\Sigma$$, consider the following problem:

\begin{aligned} & \underset{x}{\text{minimize}} & & x^T \Sigma x \\ & \text{subject to} & & x \geq 0, \; \mathbf{1}^T x = 1 \end{aligned}

Let

$\Sigma = \left[ \begin{array}{cc} a & c \\ c & b \end{array} \right]$

and note that the constraints imply that $$x_2 = 1 - x_1$$. Therefore,

\begin{aligned} x^T \Sigma x &= \left[ \begin{array}{cc} x_1 & 1 - x_1 \end{array} \right] \left[ \begin{array}{cc} a & c \\ c & b \end{array} \right] \left[ \begin{array}{c} x_1 \\ 1 - x_1 \end{array} \right] \\ \\ &= a x_1^2 + 2 c x_1 (1-x_1) + b (1 - x_1)^2 \end{aligned}

Minimizing with respect to $$x_1$$, we find that

$x_1^\text{min} = \frac{b - c}{a + b - 2c}$

and substituting $$x_1^\text{min}$$ for $$x_1$$, we obtain

$\underset{x} \min ~~ x^T \Sigma x = \frac{ab - c^2}{a + b - 2c}$

Now define the operator that takes the sum of the anti-diagonal elements of a symmetric matrix by

$\tilde{\text{tr}} (A) := \sum_{i=1}^n a_{\tilde{i} i}, ~~~~~~~~~~ \tilde{i} = n - i + 1$

and note that the minimal value can be rewritten as

$\frac{\text{det}(\Sigma)}{\text{tr}(\Sigma) - \tilde{\text{tr}}(\Sigma)}$