Recipe

Classify a real quadratic form by its signature

Use when: you have a real quadratic form, diagonal or not, and need to know how many positive, negative, and zero squares it reduces to, plus its standard classification

Goal: the signature $(p,q,r)$ and the matching name: positive definite, negative definite, indefinite, or degenerate

Taught in Signature and the classification of real quadratic forms

Quick recipe

Diagonalize the form if it isn't already
Count the positive, negative, and zero coefficients
Read off the classification from the counts
Exhibit congruence to the canonical form, if asked

1
If $Q$ has cross terms, complete the square (or use section 6's algorithm) until it's a sum of coefficient-times-square terms with no cross terms left.

Example

$Q(x,y)=x^2-4xy+3y^2$ : complete the square on the $x$ -terms, $x^2-4xy=(x-2y)^2-4y^2$ , so $Q=(x-2y)^2-4y^2+3y^2=(x-2y)^2-y^2$ .

Why: signature is read directly off a diagonal form's coefficients, so a form with cross terms has to be diagonalized first.
2
Once diagonal, count how many squared terms have a positive coefficient ( $p$ ), a negative coefficient ( $q$ ), and a zero coefficient ( $r$ ).

Example

$Q=(x-2y)^2-y^2$ has one positive coefficient and one negative coefficient, so $p=1$ , $q=1$ , $r=0$ .

Why: Sylvester's law of inertia guarantees these three counts are the same no matter which diagonalization is chosen, so they're a genuine property of $Q$ itself.
3
Match $(p,q,r)$ against the four standard names.

Example

$(p,q,r)=(1,1,0)$ has $r=0$ with both $p,q>0$ , so $Q=(x-2y)^2-y^2$ is indefinite. Compare $(p,q,r)=(1,0,1)$ for $Q(x,y)=x^2$ : $r>0$ , so this form is degenerate even though its one nonzero coefficient is positive.
- $r=0$ and $q=0$ — positive definite.
- $r=0$ and $p=0$ — negative definite.
- $r=0$ with both $p>0$ and $q>0$ — indefinite.
- $r>0$ — degenerate, regardless of $p$ and $q$ .
Why: these four names are exactly the possible shapes signature can take once $r$ and the signs of $p,q$ are pinned down, so every form falls into exactly one.
4
Rescale each nonzero diagonal entry by dividing its variable by the square root of its absolute value, turning every nonzero coefficient into $\pm1$ .

Example

For $A=\operatorname{diag}(2,3)$ , rescaling with $P=\operatorname{diag}(1/\sqrt2,1/\sqrt3)$ gives $P^{\mathsf T}AP=\operatorname{diag}(1,1)$ .

Why: two forms sharing a signature are congruent precisely because each is congruent to the same canonical $\pm1$ -and- $0$ form, and congruence is transitive.

Common mistakes

Reading the signature off a form's coefficients before it's actually been diagonalized, cross terms and all.
Miscounting a zero coefficient as either positive or negative instead of tallying it into $r$ .
Concluding "degenerate" only means "all zero," when even one zero coefficient among otherwise-nonzero ones already makes $r>0$ .
Forgetting that indefinite requires BOTH $p>0$ and $q>0$ ; a form with $p=0$ and $q>0$ is negative definite, not indefinite.