Quadratic forms, positive definiteness, Sylvester's law

In one sentence

A quadratic form is classified by its signature and by nothing else, congruence is the relation that preserves it, and congruence is not similarity — it changes eigenvalues freely.

Why the exam asks it

Because candidates arrive knowing one criterion — leading principal minors — and apply it to positive semidefiniteness, where it is false. And because congruence and similarity are easy to conflate when both are written $P^{\bullet}AP$, differing only in whether the left factor is $P^{T}$ or $P^{-1}$.

The idea

A quadratic form on $\mathbb{R}^n$ is $q(x) = x^{T}Ax$ with $A$ symmetric. The symmetry is a normalisation, not a restriction: any $A$ can be replaced by $\tfrac12(A + A^T)$ without changing $q$.

Under a change of variable $x = Py$ with $P$ invertible,

$$q(x) = (Py)^{T}A(Py) = y^{T}(P^{T}AP)y,$$

so the matrix transforms by $A \mapsto P^{T}AP$. That relation is congruence.

Congruence is $P^{T}AP$. Similarity is $P^{-1}AP$. They agree exactly when $P$ is orthogonal.

That last remark is why the spectral theorem is so useful here: for symmetric $A$ we can take $P$ orthogonal, and then congruence and similarity coincide, so the diagonal entries can be taken to be the eigenvalues.

Sylvester's law of inertia

Every real quadratic form is congruent to a diagonal form with entries $+1$, $-1$ and $0$, and the counts $(p, n, z)$ of each are invariants — independent of how you diagonalise it.

• $p$ = number of positive eigenvalues, $n$ = negative, $z$ = zero.
• $\operatorname{rank} = p + n$, signature commonly means $p - n$ or the pair $(p,n)$.

Two real symmetric matrices are congruent $\iff$ they have the same $(p, n, z)$. Note what this does not say: nothing about the eigenvalues themselves. Congruence can scale them arbitrarily, as long as no sign changes.

Over $\mathbb{C}$ there are no signs to preserve — $-1$ is a square — so only the rank classifies, and every complex quadratic form of rank $r$ is congruent to $y_1^2 + \cdots + y_r^2$.

Definiteness

$A$ symmetric is:

• positive definite if $x^TAx > 0$ for all $x \ne 0$ $\iff$ all eigenvalues $> 0$;
• positive semidefinite if $x^TAx \ge 0$ $\iff$ all eigenvalues $\ge 0$;
• indefinite if it takes both signs $\iff$ eigenvalues of both signs.

Sylvester's criterion: $A$ is positive definite $\iff$ all $n$ leading principal minors are $> 0$. This is a genuine shortcut — no eigenvalues required.

And it does not adapt to the semidefinite case. The correct statement there needs all principal minors $\ge 0$, not just the leading ones. The counterexample is small and worth knowing cold:

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

Its leading principal minors are $\det(0) = 0$ and $\det A = 0$, both $\ge 0$. Its eigenvalues are $0$ and $-1$, so it is not positive semidefinite — indeed $q(0,1) = -1$. The non-leading principal minor, the $(2,2)$ entry $-1$, is the one that catches it, and the leading sequence never looks at it.

Where intuition breaks

"Leading principal minors $\ge 0$, so positive semidefinite." $\operatorname{diag}(0,-1)$. The single most productive trap in this topic, because the definite version is true with strict inequalities and the weakening looks harmless.

"Congruent, so similar." $I$ and $\operatorname{diag}(4,1)$ are congruent via $P = \operatorname{diag}(2,1)$, and their eigenvalues are $\{1,1\}$ and $\{4,1\}$, so they are not similar. Congruence preserves only the signs.

"Similar, so congruent." Also false in general, for real symmetric matrices it happens to follow from equal eigenvalues, but similar non-symmetric matrices need not be congruent at all. Do not treat either as implying the other.

"All diagonal entries positive, so positive definite." $\begin{pmatrix}1&2\\2&1\end{pmatrix}$ has positive diagonal and eigenvalues $3$ and $-1$. The diagonal entries are $q(e_i)$ — they only test the coordinate directions.

"Positive definite, so all entries positive." $\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ is positive definite with a negative entry. The two statements are unrelated in both directions.

"$\det A > 0$, so positive definite." $-I$ in two variables has determinant $1$. The determinant is the product of the eigenvalues, and an even number of negatives cancels.

"The signature depends on how you diagonalise." It does not — that is exactly Sylvester's law, and it is the reason signature is a legitimate invariant to ask about.

"A quadratic form is determined by its matrix." By its symmetric matrix. $\begin{pmatrix}0&2\\0&0\end{pmatrix}$ and $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ give the same form $2xy$.

The exam's angle

1. Get the signature and stop. Almost every question about congruence, classification, or the shape of a conic is answered by $(p, n, z)$.
2. For definiteness, use leading minors — but only for the strict case. If the question says semidefinite, go to eigenvalues or check all principal minors.
3. Read $P^TAP$ against $P^{-1}AP$ carefully. The exponent is the whole question.
4. Count sign changes in the characteristic polynomial if the eigenvalues are awkward: Descartes' rule gives the number of positive roots quickly for a symmetric matrix.
5. Over $\mathbb{C}$, only rank matters. If the stem moves to $\mathbb{C}$, every signature distinction collapses.

The night before

• $q(x) = x^TAx$ with $A$ symmetric; change of variable gives congruence $P^TAP$.
• Congruence $\ne$ similarity. They coincide only for orthogonal $P$.
• Sylvester's law: $(p, n, z)$ is invariant. Congruent $\iff$ same $(p,n,z)$. Over $\mathbb{C}$, rank alone.
• Positive definite $\iff$ all eigenvalues $>0$ $\iff$ all leading principal minors $>0$.
• Semidefinite needs all principal minors $\ge 0$. $\operatorname{diag}(0,-1)$ has both leading minors $0$ and is not semidefinite.
• Positive diagonal does not give definiteness: $\begin{pmatrix}1&2\\2&1\end{pmatrix}$. $\det > 0$ does not either: $-I$.
• Positive definite matrices can have negative entries.
• Congruence preserves signs of eigenvalues, not their values: $I$ and $\operatorname{diag}(4,1)$.