Mathematical Methods for Economics · Linear Algebra

Quadratic forms & definiteness

A quadratic form measures the curvature of a function in every direction at once. Whether it is always positive, always negative, or a mix, decides whether an objective function is convex or concave — and hence whether a critical point is a minimum, a maximum, or a saddle. This page defines quadratic forms, extracts their symmetric matrix, and gives the two standard definiteness tests: eigenvalues and principal minors.

This page ties together the whole linear algebra sequence: it uses the determinants and the eigenvalues developed earlier. It belongs to the linear algebra part of a mathematical economics course (module code EC3304).

What is a quadratic form?

Definition — quadratic form

A quadratic form $Q$ of an $n$-vector $\mathbf{x}$ can be written

$$Q(\mathbf{x}) := \sum_{i=1}^{n}\sum_{j=1}^{n} a_{ij}x_i x_j = \mathbf{x}^{\top}A\mathbf{x},$$

for some symmetric matrix $A$, the matrix associated with $Q$.

Every term is of degree two — a square $x_i^2$ or a cross-product $x_i x_j$. Writing the form as $\mathbf{x}^{\top}A\mathbf{x}$ with $A$ symmetric is what lets us bring the full linear-algebra toolkit to bear on questions about its sign.

Extracting the associated symmetric matrix

The diagonal entry $a_{ii}$ is the coefficient of $x_i^2$; the coefficient of a cross term $x_ix_j$ is split equally between $a_{ij}$ and $a_{ji}$, which keeps $A$ symmetric.

Worked example — finding the associated matrix

Take

$$Q(\mathbf{x}) = 3x_1^2 + 6x_1x_3 + x_2^2 - 4x_2x_3 + 8x_3^2.$$

Rewrite it with each cross term shared symmetrically:

$$Q(\mathbf{x}) = 3x_1^2 + 0\,x_1x_2 + 3x_1x_3 + 0\,x_2x_1 + x_2^2 - 2x_2x_3 + 3x_3x_1 - 2x_3x_2 + 8x_3^2.$$

The coefficient $6$ of $x_1x_3$ splits into $a_{13}=a_{31}=3$; the coefficient $-4$ of $x_2x_3$ splits into $a_{23}=a_{32}=-2$. Hence

$$A=\begin{pmatrix}3&0&3\\0&1&-2\\3&-2&8\end{pmatrix}.$$

Definiteness

Definition — definiteness

A quadratic form $Q$, with symmetric matrix $A$, is

positive definite if $Q(\mathbf{x})>0$ for all $\mathbf{x}\neq\mathbf{0}$;
positive semidefinite if $Q(\mathbf{x})\ge 0$ for all $\mathbf{x}\neq\mathbf{0}$;
negative definite if $Q(\mathbf{x})<0$ for all $\mathbf{x}\neq\mathbf{0}$;
negative semidefinite if $Q(\mathbf{x})\le 0$ for all $\mathbf{x}\neq\mathbf{0}$.

$Q$ (and $A$) is indefinite if $Q(\mathbf{x})>0$ for some $\mathbf{x}$ and $Q(\mathbf{y})<0$ for some $\mathbf{y}$.

There are two standard ways to determine definiteness: using the eigenvalues of $A$, or using its principal minors. Principal minors are often preferred, especially when the eigenvalues cannot be solved for explicitly.

Test 1 — eigenvalues

Theorem — definiteness by eigenvalues

Let $A$ be a symmetric matrix (so its eigenvalues are real). Then:

$A$ is positive definite $\iff$ all eigenvalues are strictly positive;
$A$ is positive semidefinite $\iff$ all eigenvalues are nonnegative;
$A$ is negative definite $\iff$ all eigenvalues are strictly negative;
$A$ is negative semidefinite $\iff$ all eigenvalues are nonpositive;
$A$ is indefinite $\iff$ both positive and negative eigenvalues exist.

The reason is diagonalisation. Writing $A=PDP^{\top}$ and substituting $\mathbf{y}\equiv P^{\top}\mathbf{x}$,

$$\mathbf{x}^{\top}A\mathbf{x} = \mathbf{x}^{\top}(PDP^{\top})\mathbf{x} = \mathbf{y}^{\top}D\mathbf{y} = \sum_{i=1}^{n}\lambda_i y_i^2.$$

The form is a weighted sum of squares with the eigenvalues as weights, so its sign is dictated entirely by the signs of the $\lambda_i$.

Worked example — definiteness from eigenvalues

Classify $Q(\mathbf{x}) = -x_1^2 + 6x_1x_2 - 9x_2^2 - 2x_3^2$. The associated matrix is

$$A=\begin{pmatrix}-1&3&0\\3&-9&0\\0&0&-2\end{pmatrix}.$$

Its characteristic equation factorises as $\lvert A-\lambda I\rvert = -\lambda(\lambda+2)(\lambda+10) = 0$, giving eigenvalues $0, -2, -10$. All are nonpositive (with one exactly zero), so $A$ is negative semidefinite.

Principal minors

Definition — principal and leading principal minors

Principal minors of $A$ are the determinants of submatrices obtained by deleting the same rows and columns of $A$.

Leading principal minors are obtained by deleting the last row(s) and column(s) — the top-left $1\times1, 2\times2, \ldots$ blocks. A (leading) principal minor is of order $r$ if its submatrix is $r\times r$; the whole matrix $A$ is itself the leading principal minor of order $n$.

Write $D_k^\ast$ for the leading principal minor of order $k$, and $D_k$ for an arbitrary principal minor of order $k$.

Test 2 — principal minors

Theorem — definiteness by principal minors

$A$ is positive definite $\iff D_k^\ast > 0$ for all $k$;
$A$ is positive semidefinite $\iff$ every $D_k \ge 0$ for all $k$;
$A$ is negative definite $\iff (-1)^k D_k^\ast > 0$ for all $k$ (leading minors alternate in sign, starting negative);
$A$ is negative semidefinite $\iff$ every $(-1)^k D_k \ge 0$ for all $k$.

Note the asymmetry: for the definite cases the cheap leading minors $D_k^\ast$ suffice, but the semidefinite cases require checking all principal minors $D_k$.

Worked example — definiteness from leading principal minors

Determine the definiteness of the matrix from the earlier example,

$$A=\begin{pmatrix}3&0&3\\0&1&-2\\3&-2&8\end{pmatrix}.$$

The leading principal minors are

$$D_1^\ast = 3, \qquad D_2^\ast = \begin{vmatrix}3&0\\0&1\end{vmatrix} = 3, \qquad D_3^\ast = \lvert A\rvert = 3.$$

(For $D_3^\ast$: $3(1\cdot 8 - (-2)(-2)) - 0 + 3(0\cdot(-2) - 1\cdot 3) = 3(4) + 3(-3) = 3$.) All three are strictly positive, so $A$ is positive definite.

Why definiteness matters in economics

Concavity, convexity and optimisation

The notion of definiteness plays a central role in determining the concavity or convexity of a multivariate function. A twice-differentiable function is convex on a region when its Hessian (the symmetric matrix of second partial derivatives) is positive semidefinite there, and concave when the Hessian is negative semidefinite. Definiteness is therefore exactly the tool that classifies critical points and settles the second-order conditions of an optimisation problem: a positive definite Hessian at a stationary point signals a local minimum, a negative definite Hessian a local maximum, and an indefinite Hessian a saddle point.

Because the Hessian is symmetric, its principal minors can be checked directly. For example, to show a function is convex it is enough to verify that every principal minor of its Hessian is nonnegative throughout the domain — a pure linear-algebra calculation applied to a matrix whose entries happen to be second derivatives. This is why the two tests on this page reappear constantly in consumer theory, producer theory and any problem that maximises or minimises a smooth objective.

Self-check questions

1. Write the quadratic form $Q(\mathbf{x}) = 2x_1^2 + 4x_1x_2 + 5x_2^2$ as $\mathbf{x}^\top A\mathbf{x}$, and classify its definiteness using leading principal minors.

Split the cross term $4x_1x_2$ equally: $a_{12}=a_{21}=2$. So

$$A=\begin{pmatrix}2&2\\2&5\end{pmatrix}.$$

Leading principal minors: $D_1^\ast = 2 > 0$ and $D_2^\ast = \lvert A\rvert = 2(5)-2(2) = 6 > 0$. Both positive, so $A$ is positive definite — $Q(\mathbf{x})>0$ for every $\mathbf{x}\neq\mathbf{0}$.

2. A symmetric $2\times2$ matrix has eigenvalues $4$ and $-1$. What is its definiteness, and what would that imply if it were the Hessian of an objective at a critical point?

The eigenvalues have opposite signs (one positive, one negative), so the matrix is indefinite.

If this were the Hessian at a stationary point of a function, the point would be a saddle point — a maximum in one direction and a minimum in another — so it is neither a local maximum nor a local minimum.

3. Classify $Q(\mathbf{x}) = -2x_1^2 - 3x_2^2$ by inspection, and confirm with the negative-definite minor test.

By inspection $Q$ is a negative-weighted sum of squares, so $Q(\mathbf{x})<0$ for all $\mathbf{x}\neq\mathbf{0}$: negative definite.

Matrix $A=\begin{pmatrix}-2&0\\0&-3\end{pmatrix}$. The negative-definite condition is $(-1)^k D_k^\ast > 0$: for $k=1$, $(-1)^1 D_1^\ast = -(-2) = 2 > 0$ ✓; for $k=2$, $(-1)^2 D_2^\ast = \lvert A\rvert = 6 > 0$ ✓. The leading minors alternate in sign as required, confirming negative definiteness.

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Eigenvalues & diagonalisation

Where the eigenvalue test comes from — diagonalising a symmetric matrix as a sum of squares.

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Matrices & linear systems

Determinants — the computation behind every principal minor.

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Worked examples & practice problems

Full solutions to a parametric linear system, a matrix inversion and an eigenvalue problem.

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