Principal Minors and Sylvester's Criterion¶

Theorem¶

Let $\boldsymbol{A}\in\mathbb{R}^{n\times n}$ and let $I=\{i_1,\ldots,i_k\}\subseteq\{1,\ldots,n\}$$\{1,\dots ,n\}$. The principal submatrix $[\boldsymbol{A}]_I$ keeps the rows and columns indexed by $I$, and its principal minor is $\det([\boldsymbol{A}]_I)$. The $k$-th leading principal submatrix is $\boldsymbol{A}_k=[\boldsymbol{A}]_{\{1,\ldots,k\}}$$[\mathbf{A}{]}_{\{1,\dots ,k\}}$, and its determinant $\det(\boldsymbol{A}_k)$ is the $k$-th leading principal minor.

If $\boldsymbol{A}$ is symmetric, Sylvester's criterion states:

• $\boldsymbol{A}$ is positive definite if and only if $\det(\boldsymbol{A}_k)>0$ for all $k=1,\ldots,n$$1,\dots ,n$.
• $\boldsymbol{A}$ is negative definite if and only if $(-1)^k\det(\boldsymbol{A}_k)>0$ for all $k=1,\ldots,n$$1,\dots ,n$.
• Positive and negative semidefinite cases require all principal minors with the corresponding sign pattern.

Interpretation¶

Principal minors are determinants of square submatrices that keep the same rows as columns, unlike general minors. Sylvester's criterion reads the definiteness of a symmetric matrix from the signs of its $n$ leading principal minors, usually quicker than computing eigenvalues.

The nested boxes are the leading principal submatrices $\boldsymbol{A}_1$, $\boldsymbol{A}_2$ and $\boldsymbol{A}_3=\boldsymbol{A}$ of a $3\times3$ matrix: each is the top-left $k\times k$ block, and $\det\boldsymbol{A}_1=a_{11}$$a_{11}$. Positive definiteness requires the signs $+,+,+$ for their determinants, negative definiteness the alternating signs $-,+,-$.

If $\boldsymbol{A}$ is positive definite, so is each $\boldsymbol{A}_k$: for $\boldsymbol{x}$ with only its first $k$ entries nonzero, $\boldsymbol{x}^{\top}\boldsymbol{A}\boldsymbol{x}$ is the quadratic form of $\boldsymbol{A}_k$. A positive definite matrix has positive eigenvalues (definiteness and eigenvalues) and its determinant is their product, so every $\det\boldsymbol{A}_k>0$$0$. For negative definiteness, apply this to $-\boldsymbol{A}$ and use $\det(-\boldsymbol{A}_k)=(-1)^k\det\boldsymbol{A}_k$$(-1{)}^k\det {\mathbf{A}}_k$: that is where the alternation comes from.

Symmetry matters: the non-symmetric matrix $\begin{pmatrix}1&-3\\0&1\end{pmatrix}$ has leading minors $1$ and $1$, yet $\boldsymbol{x}^{\top}\boldsymbol{A}\boldsymbol{x}=-1$ at $\boldsymbol{x}=(1,1)$$(1,1)$. For semidefiniteness the leading minors are not enough: $\operatorname{diag}(0,-1)$ has leading minors $0$ and $0$, and only the principal minor $-1$ from row and column $2$ shows that it is not positive semidefinite.

A common slip is to read "negative definite" as "all leading minors negative". For $\operatorname{diag}(-2,-3)$ from the Examples the leading minors are $-2$ and $6$: the even-order ones are positive. The criterion is applied to Hessians, and in bordered form to constrained directions.

Key Results¶

• Leading principal minors are enough for positive and negative definiteness.
• Semidefinite cases require checking all principal minors.
• The criterion applies to symmetric matrices.

Examples¶

• For $\boldsymbol{A}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$$\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, the leading minors are $2$ and $3$, so $\boldsymbol{A}$ is positive definite.
• For $\boldsymbol{A}=\begin{pmatrix}-2&0\\0&-3\end{pmatrix}$$\left(\begin{array}{cc}-2& 0\\ 0& -3\end{array}\right)$, the alternating sign test shows negative definiteness.

Prerequisites¶

• Matrix
• Definite Matrix

Related Concepts¶

• Determinant
• Bordered Matrix Test

Sources¶

• Math IV: Multivariate Calculus · Week 3, Lecture 2

Aliases: Sylvester criterion Keywords: principal minor, leading principal minor, Sylvester's criterion, principal submatrix Symbols: element of, real numbers, subset

Last updated: 20 October 2026 03:02