What is quadratic form? — Mathematics

Definition 12.1 University Mathematics — Year 2 · Chapter 12 — Quadratic Forms

A symmetric bilinear form on a real vector space $E$ is a bilinear $\varphi \colon E \times E \to \R$ with $\varphi(x, y) = \varphi(y, x)$ ; the associated quadratic form is $q(x) = \varphi(x, x)$ . The form $\varphi$ is recovered from $q$ by polarization:

\varphi(x, y) = \tfrac12\bigl(q(x + y) - q(x) - q(y)\bigr).

In a basis $(e_i)$ , the matrix of $\varphi$ is the symmetric $B = (\varphi(e_i, e_j))$ , with $q(x) = X^{\mathsf T} B X$ ; a change of basis with matrix $P$ replaces $B$ by $P^{\mathsf T} B P$ (congruence — not similarity!). The rank of $q$ is $\operatorname{rk} B$ (invariant: congruence multiplies by invertible matrices).

Examples

Example 12.3 (Congruence in action)

Take $q(x, y) = x^2 + 4xy + y^2$ (matrix $B = \begin{pmatrix} 1 & 2\\ 2 & 1\end{pmatrix}$ ) and the new basis $e_1' = (1, 1)$ , $e_2' = (1, -1)$ , i.e. $P = \begin{pmatrix} 1 & 1\\ 1 & -1\end{pmatrix}$ . Then

P^{\mathsf T}BP = \begin{pmatrix} 1 & 1\\ 1 & -1\end{pmatrix} \begin{pmatrix} 1 & 2\\ 2 & 1\end{pmatrix} \begin{pmatrix} 1 & 1\\ 1 & -1\end{pmatrix} = \begin{pmatrix} 6 & 0\\ 0 & -2\end{pmatrix} :

in the coordinates $(u, v)$ along the new basis, $q = 6u^2 - 2v^2$ — check: $x = u + v$ , $y = u - v$ gives $x^2 + 4xy + y^2 = 6u^2 - 2v^2$ directly. Note the new diagonal entries $6, -2$ are not the eigenvalues $3, -1$ of $B$ : congruence rescales, only similarity preserves spectra — but the signs agree, as Sylvester’s theorem demands. (The basis here is orthogonal but not orthonormal; normalizing it by $\frac{1}{\sqrt2}$ would divide the diagonal by $2$ and recover the eigenvalues.)

Example 12.4 (Gram determinants measure area)

For $v_1, v_2$ in a Euclidean space, the Gram matrix $G = \bigl(\langle v_i, v_j\rangle\bigr)$ packages lengths and angle; its determinant packages area:

\det G = \norm{v_1}^2\norm{v_2}^2 - \langle v_1, v_2\rangle^2 = \norm{v_1}^2\norm{v_2}^2\bigl(1 - \cos^2\theta\bigr) = \bigl(\norm{v_1}\,\norm{v_2}\sin\theta\bigr)^2 ,

the squared area of the parallelogram on $v_1, v_2$ — and Cauchy–Schwarz is exactly the statement $\det G \geq 0$ . Worked instance: $v_1 = (1, 2, 2)$ , $v_2 = (2, 1, -2)$ in $\R^3$ :

G = \begin{pmatrix} 9 & 0\\ 0 & 9 \end{pmatrix}, \qquad \det G = 81 :

the vectors are orthogonal of length $3$ , spanning a parallelogram (here, a square) of area $\sqrt{81} = 9$ . Closing insight: no cross product and no dimension- $3$ magic was used — $\sqrt{\det G}$ measures $k$ -dimensional volume in any dimension, the starting point of the weekend problem’s Part I and of the surface-area integrals later in this volume.

Example 12.16

$A = \begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}$ : eigenvalues $3$ (eigenvector $\frac{1}{\sqrt2}(1,1)$ ) and $1$ ( $\frac{1}{\sqrt2}(1,-1)$ ). The quadratic form $2x^2 + 2xy + 2y^2$ becomes $3X^2 + Y^2$ in the rotated orthonormal frame: an ellipse’s principal axes, computed. The Gauss reduction reaches a diagonal form too, but only the spectral theorem reaches it without distorting lengths.

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