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

\[\frac{\text{det}(\Sigma)}{\text{tr}(\Sigma) - \tilde{\text{tr}}(\Sigma)}\] Written on November 6th, 2018 by Jordan Bryan