quadratic form কী?

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

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:

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 $\R^3$:

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.