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TL;DR

A symmetric matrix A is positive definite if \( \mathbf{x}^T A \mathbf{x} > 0 \) for all non-zero \( \mathbf{x} \). Equivalently, all eigenvalues are strictly positive.

By Valenke Exam Prep Team·Last updated 2026-06-03

Positive Definite Matrix

A symmetric matrix A is positive definite if \( \mathbf{x}^T A \mathbf{x} > 0 \) for all non-zero \( \mathbf{x} \). Equivalently, all eigenvalues are strictly positive.

Why it matters for interviews

Covariance matrices must be positive semi-definite. Portfolio optimization requires this property -- without it, the optimization problem is ill-defined. Understanding when estimated covariance matrices lose positive definiteness is practically important.

Definition and Mathematical Foundation

A symmetric matrix A is positive definite if \( \mathbf{x}^T A \mathbf{x} > 0 \) for all non-zero \( \mathbf{x} \). Equivalently, all eigenvalues are strictly positive.

Application in Quantitative Finance

Covariance matrices must be positive semi-definite. Portfolio optimization requires this property -- without it, the optimization problem is ill-defined. Understanding when estimated covariance matrices lose positive definiteness is practically important.

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Frequently Asked Questions

Why must covariance matrices be positive semi-definite?▼
Because \( \text{Var}(\mathbf{a}^T \mathbf{X}) = \mathbf{a}^T \Sigma \mathbf{a} \geq 0 \) for all vectors a. Variance cannot be negative, which is exactly the PSD condition.
When can estimated covariance matrices fail to be positive definite?▼
When the number of observations is less than the number of variables (p > n). The sample covariance matrix becomes singular. Shrinkage estimators or factor models restore positive definiteness.
What is Cholesky decomposition and why does it require PD?▼
Cholesky factors \( A = LL^T \) where L is lower triangular. It requires positive definiteness because the algorithm involves square roots of diagonal elements, which must be positive.