ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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$q (x, y) = 2 x^{2} + 4 x y + 5 y^{2} = 2 (x + y)^{2} + 3 y^{2}$ has inertia $(2, 0, 0)$

Example

The real quadratic form

$q (x, y) = 2 x^{2} + 4 x y + 5 y^{2}$

satisfies $q (x, y) = 2 (x + y)^{2} + 3 y^{2}$ and has inertia $(2, 0, 0)$ .

Facts & Assumptions

Given: The displayed real quadratic form, whose symmetric matrix is $A = (\begin{array}{cc} 2 & 2 \\ 2 & 5 \end{array})$ .

[L1]

Quadratic forms over characteristic not $2$ admit diagonal coordinates (Over a field of characteristic not $2$ , every quadratic form has diagonal coordinates $q (x) = a_{1} x_{1}^{2} + \dots + a_{n} x_{n}^{2}$ ).

[L2]

The counts of positive, negative, and zero diagonal entries give the unique inertia (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

Verification

technique · explicit congruence and expansion

1.1

Expanding gives $2 (x + y)^{2} + 3 y^{2} = 2 x^{2} + 4 x y + 5 y^{2} = q (x, y)$ . With $u = x + y$ , $v = y$ , old coordinates are $(x, y) = (u - v, v)$ , so $P = (\begin{array}{cc} 1 & - 1 \\ 0 & 1 \end{array})$ is invertible.

algebra

2.1

Direct multiplication gives $P^{T} A P = (\begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array})$ . Both diagonal entries are positive, and positive rescaling changes this matrix to $I_{2}$ .

step 1.1algebra

3.1

By [L2], the inertia is $(2, 0, 0)$ , in agreement with the diagonalization promised by [L1].

step 2.1L1L2∎

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q(x,y)=2x2+4xy+5y2=2(x+y)2+3y2q(x,y)=2x2+4xy+5y2=2(x+y)2+3y2 has inertia (2,0,0)(2,0,0)