7. Optional supplement: quadratic forms

This supplement is optional; its inclusion does not establish any current course or examination requirement. Work over $\mathbb R$.

Definition

Quadratic form

A quadratic form on $\mathbb R^n$ is $q(x)=x^{\mathsf T}Ax$ with $A$ real symmetric. It is positive definite if $q(x)>0$ for every nonzero $x$, positive semidefinite if $q(x)\ge0$ for all $x$, and indefinite if it takes both positive and negative values.

Every real matrix $B$ gives the same form as its symmetric part $(B+B^{\mathsf T})/2$. The symmetric representation is unique. Changing coordinates by $x=Py$ gives $q(x)=y^{\mathsf T}P^{\mathsf T}APy$; this is congruence, not similarity in general. If $P=Q$ is orthogonal, however, the two formulas coincide.

The preceding real spectral theorem gives an orthogonal $Q$ with $Q^{\mathsf T}AQ=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Substitution $x=Qy$ gives $q(Qy)=\sum_i\lambda_i y_i^2$, providing the diagonal representation used in the shared inertia proof. Testing coordinate vectors gives necessity of the eigenvalue sign conditions; sums of squares give sufficiency.

All spaces are finite-dimensional over the stated field. A bilinear form is linear in both arguments; it is not a Hermitian inner product over $\mathbb C$.

Rescaling the diagonal entries of a real symmetric form gives

$$q= x_1^2+\cdots+x_{p_+}^2-y_1^2-\cdots-y_{p_-}^2+0\cdot z_1^2+\cdots+0\cdot z_{p_0}^2.$$

The triple $(p_+,p_-,p_0)$ is its inertia and is independent of all basis choices.

Proof of uniqueness. In this representation, the positive coordinate subspace has dimension $p_+$ and the quadratic form $q$ is positive definite on it. No subspace of dimension greater than $p_+$ can carry a positive definite restriction: it must intersect the coordinate subspace where all positive coordinates vanish, whose dimension is $p_-+p_0$, because their dimensions sum to more than $n$. A nonzero vector in that intersection has quadratic value at most zero, a contradiction. Thus $p_+$ is intrinsically the largest possible dimension of a positive definite subspace. Similarly $p_-$ is the largest dimension for a negative definite restriction. Then $p_0=n-p_+-p_-$, equivalently the radical dimension. ∎

A form is positive definite exactly when $(p_+,p_-,p_0)=(n,0,0)$, negative definite when $(0,n,0)$, and indefinite when both $p_+,p_->0$. Positive semidefinite allows zeros but no negative squares. Nondegeneracy means $p_0=0$, not positivity.

For a real symmetric matrix, orthogonal spectral diagonalization is a particular congruence since $Q^{-1}=Q^{\mathsf T}$. It follows that inertia counts positive, negative and zero eigenvalues with multiplicity. General congruence preserves those counts, not the numerical eigenvalues themselves.

Worked example

Rotate to principal axes

For $q(x,y)=2x^2+2xy+2y^2$, $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$. With $u=(x+y)/\sqrt2$ and $v=(x-y)/\sqrt2$, spectral decomposition gives $q=3u^2+v^2$. Both eigenvalues are positive, so the form is positive definite and its inertia is $(2,0,0)$, counting positive, negative and zero signs.

Choosing a method and avoiding mistakes

Use orthogonal diagonalization to find principal axes and eigenvalues. Completing squares may classify signs more quickly, but the resulting invertible change need not be orthogonal. A cross-term coefficient $2bxy$ corresponds to off-diagonal entries $b$, not $2b$. Positive determinant alone does not imply positive definiteness.

Exercises

1. Classify $q(x,y)=x^2-4xy+y^2$ by its eigenvalue signs.
2. Classify $q(x,y)=x^2+2xy+y^2$ and find its zero set.
3. Explain why a nonsingular coordinate change cannot turn $x^2-y^2$ into a positive-definite form.

Full solutions

1. The symmetric matrix is $\begin{pmatrix}1&-2\\-2&1\end{pmatrix}$, with eigenvalues $3$ and $-1$. Therefore the form is indefinite with inertia $(1,1,0)$.
2. $q=(x+y)^2\ge0$; it is positive semidefinite but not definite because it vanishes for nonzero $(1,-1)$. Its zero set is the line $x+y=0$.
3. For any invertible $P$, the vector $y=P^{-1}(0,1)$ is nonzero and the transformed form has value $-1$ at $y$. Thus it cannot be positive definite. Equivalently its inertia remains $(1,1,0)$.

Connection

In optimization, a Hessian at a critical point is symmetric; its quadratic form describes the second-order term. A definite Hessian supplies a local test under the usual differentiability hypotheses, whereas a semidefinite Hessian alone is inconclusive.