Convexity of -log(det(X))

• categories: Linear algebra, Optimization, Theorem

Theorem:

The function $f(X)=-\log (\det (X))$ is Convex Function over the domain of positive definite matrices, $X\in {\mathbb{S}}_{++}^n$, where ${\mathbb{S}}_{++}^n$ is the set of $n\times n$ symmetric positive definite matrices.

Proof Outline

Step 1: Domain and Feasibility

• The determinant $\det (X)$ is positive for all $X\in {\mathbb{S}}_{++}^n$.
• $\log (\det (X))$ is well-defined and finite over ${\mathbb{S}}_{++}^n$.
• $-\log (\det (X))$ is thus a valid function on ${\mathbb{S}}_{++}^n$.

Step 2: First and Second Derivatives

To show convexity, we compute the Hessian of $f(X)$ and verify that it is positive semidefinite.

1. Gradient of $f(X)$: Using matrix calculus:

   $$f(X)=-\log (\det (X))⟹\nabla f(X)=-X^{-1}.$$

2. Hessian of $f(X)$: The gradient $\nabla f(X)$ is a matrix-valued function. Differentiating $\nabla f(X)$ with respect to $X$ gives the Hessian:

   $${\nabla }^{2}f(X)[H,H]=\text{tr}(X^{-1}H{X}^{-1}H),$$

   for any symmetric perturbation matrix $H\in {\mathbb{S}}^n$.

Step 3: Positive Semidefiniteness

• The quadratic form of the Hessian is: $${\nabla }^{2}f(X)[H,H]=\text{tr}(X^{-1}H{X}^{-1}H).$$
• Since $X^{-1}$ is positive definite (as $X\in {\mathbb{S}}_{++}^n$) and $H$ is symmetric, the term $X^{-1}H$ is also symmetric. The trace of a square of a symmetric matrix is non-negative: $$\text{tr}(X^{-1}H{X}^{-1}H)\ge 0.$$
• Thus, the Hessian is positive semidefinite.

Step 4: Convexity

The positive semidefiniteness of the Hessian implies that $f(X)$ is convex over ${\mathbb{S}}_{++}^n$.

Intuition

1. The determinant $\det (X)$ measures the volume of the parallelepiped defined by the rows or columns of $X$. Minimizing $-\log (\det (X))$ encourages $X$ to “spread out” the volume.

2. The function $-\log (\det (X))$ grows rapidly as $X$ approaches singularity (where $\det (X)\to 0$), discouraging near-singular configurations.