Quadratic forms and definiteness

From a polynomial to a matrix

Definition. A quadratic form in $x = (x_1, \dots, x_n)$ is a polynomial in which every term has degree exactly 2. It can always be written as

$q(x) = x^\top A x = \sum_{i,j} a_{ij} x_i x_j$

with $A$ symmetric. The squared terms go on the diagonal, and each cross term is split equally between the two off-diagonal positions.

Example 1. $q(x, y) = 2x^2 + 6xy + 5y^2$ has $A = \begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix}$. Check: $x^\top A x = 2x^2 + 3xy + 3yx + 5y^2$.

Example 2. $q = x_1^2 - 2x_2^2 + 4x_1x_3 - 6x_2x_3$ has $A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & -2 & -3 \\ 2 & -3 & 0 \end{pmatrix}.$

A non-symmetric matrix gives the same form as its symmetric part. $x^\top A x = x^\top \frac{A + A^\top}{2} x$, because the skew part contributes 0. For example, $\begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ both give $x^2 + 4xy + y^2$.

Definiteness

Name	Condition for all $x \ne 0$	Eigenvalues of $A$
Positive definite (PD)	$q(x) > 0$	all $> 0$
Positive semidefinite (PSD)	$q(x) \ge 0$	all $\ge 0$
Negative definite	$q(x) < 0$	all $< 0$
Negative semidefinite	$q(x) \le 0$	all $\le 0$
Indefinite	takes both signs	some $> 0$ and some $< 0$

Tests

1. Eigenvalues, as in the table.
2. Sylvester's criterion. $A$ is PD exactly when all leading principal minors are positive: $D_1 = a_{11} > 0$, $D_2 = \det$ of the top-left $2 \times 2$ block $> 0$, …, $D_n = \det A > 0$.
3. Negative definite exactly when the leading minors alternate in sign: $D_1 < 0$, $D_2 > 0$, $D_3 < 0$, … Equivalently, $-A$ is PD.
4. Semidefinite needs all principal minors (not only the leading ones) to be $\ge 0$.
5. 2 × 2 shortcut. For $\begin{pmatrix} a & b \\ b & c\end{pmatrix}$:
   • PD exactly when $a > 0$ and $ac - b^2 > 0$;
   • indefinite exactly when $ac - b^2 < 0$.

Quick necessary conditions.

• PD needs every diagonal entry to be positive. A zero or negative diagonal entry rules it out at once.
• PD needs $\det A > 0$ and $\operatorname{trace} A > 0$.
• These conditions are necessary but not sufficient.

Why the semidefinite case needs all minors. $\begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$ has leading minors 0 and 0, both $\ge 0$. Yet $q = -y^2 \le 0$, so it is not PSD. The principal minor $-1$ (the bottom-right entry) reveals this.

Solved examples

Example 3: a positive definite 3 × 3 matrix. $A = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{pmatrix}$.

• $D_1 = 2$ and $D_2 = 4 - 1 = 3$.
• $D_3 = 2(4 - 1) - (-1)(-2 - 0) + 0 = 6 - 2 = 4$.

All are positive, so $A$ is PD. Its eigenvalues are $2 - \sqrt2$, $2$ and $2 + \sqrt2$, whose product is $(4 - 2) \times 2 = 4 = D_3$.