Bordered Matrix Test¶

Theorem¶

Let $Q(\boldsymbol{x})=\boldsymbol{x}^{\top} \boldsymbol{A}\boldsymbol{x}$ with $\boldsymbol{A}\in\mathbb{R}^{n\times n}$ symmetric, and let the independent linear constraint set be $\boldsymbol{B}\boldsymbol{x}=0$ with $\boldsymbol{B}\in\mathbb{R}^{m\times n}$ and $\operatorname{rank}(\boldsymbol{B})=m$ $m$ . The bordered matrix is

\boldsymbol{H}= \begin{pmatrix} \boldsymbol{0}_m & \boldsymbol{B}\\ \boldsymbol{B}^{\top} & \boldsymbol{A} \end{pmatrix}.

Ignore the first $2m$ leading principal minors. For $k=2m+1,\ldots,m+n$ $n$ :

If $(-1)^m\det(\boldsymbol{H}_k)>0$ for all relevant $k$ , then $Q$ is positive definite on the constraint set.
If $(-1)^{k-m}\det(\boldsymbol{H}_k)>0$ for all relevant $k$ , then $Q$ is negative definite on the constraint set.

Interpretation¶

The bordered matrix test checks the sign of a quadratic form only along the directions that linear constraints leave free (definiteness on the constraint set). It is the constrained version of Sylvester's criterion, again read from determinant signs.

Block matrix H with a grey zero block, red border blocks B and B transpose, and the blue matrix A of the quadratic form.

The figure shows the blocks of $\boldsymbol{H}$ : an $m\times m$ zero block (grey), the constraint matrix $\boldsymbol{B}$ and its transpose as a border (red), and the matrix $\boldsymbol{A}$ of the form (blue). $\boldsymbol{H}_k$ is the top-left $k\times k$ submatrix. The first $2m$ leading minors do not depend on $\boldsymbol{A}$ , which is why they are ignored; the remaining $n-m$ are as many as the dimension of the constraint set $\{\boldsymbol{x}\colon\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}\}$ $0}$ .

For $x_1^2-x_2^2$ on the line $x_1=2x_2$ from the Examples, $\boldsymbol{B}=\begin{pmatrix}1&-2\end{pmatrix}$ and $\boldsymbol{A}=\operatorname{diag}(1,-1)$ $diag (1, - 1)$ , so $\det\boldsymbol{H}=-3$ and $(-1)^1\det\boldsymbol{H}>0$ $0$ ; indeed $Q(2t,t)=3t^2>0$ for $t\neq0$ $0$ .

The rank condition says that the $m$ constraints are independent, so they remove exactly $m$ directions. Without it the test breaks down: if the rows of $\boldsymbol{B}$ are dependent, some $\boldsymbol{w}\neq\boldsymbol{0}$ has $\boldsymbol{B}^{\top}\boldsymbol{w}=\boldsymbol{0}$ $0$ , then $\boldsymbol{H}$ sends $(\boldsymbol{w},\boldsymbol{0})$ to $\boldsymbol{0}$ , and $\det\boldsymbol{H}=0$ whatever $\boldsymbol{A}$ is.

A common slip is to expect positive determinants for a positive definite form. With one constraint, positive definiteness on the constraint set shows up as negative minors, as in the example; the factor $(-1)^m$ tells which sign to look for. The test returns later in the course, with a Lagrangian's Hessian as $\boldsymbol{A}$ , as the bordered Hessian.

Key Results¶

Definiteness on the whole space can change after restricting to a constraint set.
For positive definiteness, all relevant signs match $(-1)^m$ .
For negative definiteness, relevant signs alternate and the full determinant has sign $(-1)^n$ .

Examples¶

The form $x_1^2-x_2^2$ is indefinite on $\mathbb{R}^2$ but positive on the line $x_1=2x_2$ $2 x_{2}$ .
When $n=2$ and $m=1$ $1$ , there is just one relevant minor: $\det \boldsymbol{H}<0$ gives positive definiteness of $Q$ on the constraint line.
When $n=3$ and $m=1$ $1$ , inspect $\det(\boldsymbol{H}_3)$ and $\det(\boldsymbol{H}_4)$ after ignoring the first two leading minors.

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 3, Lecture 2

Keywords: bordered matrix, bordered determinant test, constrained definiteness test, quadratic form under constraints Symbols: element of, real numbers

Last updated: 20 October 2026 03:02