SVD, QR, Cholesky and Positive Definiteness

A symmetric matrix $A$ turns a vector into a single number through its quadratic form $\mathbf{x}^T A\mathbf{x}$. Call $A$ positive definite (PD) if $\mathbf{x}^T A\mathbf{x} > 0$ for every nonzero $\mathbf{x}$, and positive semidefinite (PSD) if $\mathbf{x}^T A\mathbf{x} \ge 0$. Picture the surface $z = \mathbf{x}^T A\mathbf{x}$: a bowl for PD, a bowl with a flat valley for PSD but singular, and a saddle when $A$ has eigenvalues of both signs.

For a symmetric matrix there are two quick tests. It is PD exactly when every eigenvalue is positive, and exactly when every leading principal minor (the determinants of the top-left $1\times1$, $2\times2$, up to $n\times n$ blocks) is positive. For $A = \begin{pmatrix} 4 & 2 \\ 2 & 3 \end{pmatrix}$ the minors are $4$ and $12 - 4 = 8$, so $A$ is PD. The leading-minor shortcut only certifies strict PD; for PSD you must check every principal minor.

Quants care because every covariance matrix is PSD. For any weight vector $\mathbf{w}$,

since no portfolio has negative variance. $\Sigma$ is strictly PD unless some nonzero portfolio has zero variance, which happens exactly when one asset's return is a fixed linear combination of the others plus a constant.

Interview version: can three assets have $\rho_{12} = 0.9$, $\rho_{13} = 0.9$, $\rho_{23} = -0.5$? Each number is a legal correlation, but the determinant of the correlation matrix is

so some eigenvalue is negative and some portfolio would have negative variance. The angle picture from Vectors and Dot Products says the same thing: assets 2 and 3 both sit close to asset 1, so $\rho_{23} \ge 2(0.9)^2 - 1 = 0.62$.