Sylvester's criterion

Statement

Let $A$ be an $n \times n$ complex Hermitian matrix. For $1 \le k \le n$, let $\Delta_k$ be the determinant of the upper left $k \times k$ submatrix of $A$. Then $A$ is positive definite if and only if $\Delta_k > 0$ for all $k = 1, \dots, n$.

Remarks

The numbers $\Delta_k$ are the leading principal minors of $A$. For $\begin{pmatrix} a & b \\ b & c \end{pmatrix}$ real, the test is $a > 0$ and $ac - b^2 > 0$. For positive semidefinite matrices, the analogous test needs all principal minors, not only the leading ones, to be nonnegative.

Explanations

L1 human

Sylvester's criterion tests positive definiteness with $n$ determinants. We check the top left $1 \times 1$, $2 \times 2$, and so on, up to the full matrix. All of them must be positive.