Quadratic Forms, Signature, and Principal Axes
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with matrix multiplication, the concept of a symmetric matrix, and the basic properties of determinants. Familiarity with the Spectral Theorem for real symmetric matrices is assumed.
Why this matters
Quadratic forms encode the geometry of physical systems, from the shape of potential energy surfaces to the inertia of rotating bodies.
In classical mechanics, the kinetic and potential energies of a system of coupled oscillators are naturally expressed as quadratic forms in the generalized coordinates. Understanding how to simplify these forms by rotating the coordinate system—finding the principal axes—is essential for decoupling the equations of motion. This lecture bridges the gap between abstract linear algebra and the concrete geometric analysis of physical systems, showing how the eigenvalues of a symmetric matrix determine the "shape" of the associated quadratic surface.
The Geometry of Quadratic Forms
A quadratic form is a polynomial of degree two in several variables, with no linear or constant terms. For example, in two dimensions, an expression like \( Q(x,y) = ax^2 + bxy + cy^2 \) is a quadratic form. Geometrically, the level sets \( Q(x,y) = k \) (for constant \( k \)) describe central conics: ellipses, hyperbolas or, when the form is degenerate, pairs of parallel lines (never parabolas, because a quadratic form has no linear terms). The coefficients \( a, b, c \) determine the orientation and shape of these curves. If the form is positive definite, the level sets are ellipses centered at the origin; if indefinite, they are hyperbolas. The goal of principal axis analysis is to find a new coordinate system where the cross-term \( bxy \) vanishes, leaving only pure squares. This diagonalization reveals the principal axes of the ellipse or hyperbola, aligning the coordinate axes with the natural symmetry directions of the system.
Matrix Representation and Symmetry
Any quadratic form \( Q(\mathbf{x}) \) in \( n \) variables can be written compactly as \( Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} \), where \( \mathbf{x} \) is a column vector and \( A \) is a symmetric matrix. The symmetry of \( A \) is not a restriction but a necessity: the coefficient of \( x_i x_j \) in the expansion is \( A_{ij} + A_{ji} \), so we can always choose \( A_{ij} = A_{ji} \) without loss of generality. For instance, \( Q(x,y) = 2x^2 + 4xy + 3y^2 \) corresponds to the matrix \( A = \begin{pmatrix} 2 & 2 \\ 2 & 3 \end{pmatrix} \). The diagonal entries \( A_{ii} \) are the coefficients of the squared terms, while the off-diagonal entries \( A_{ij} \) (for \( i \neq j \)) are half the coefficients of the mixed terms \( x_i x_j \). This matrix representation allows us to apply linear algebra tools, such as eigenvalue decomposition, to analyze the form.
Signature and Classification
The signature of a quadratic form is a pair of integers \( (p, q) \) where \( p \) is the number of positive eigenvalues and \( q \) is the number of negative eigenvalues of the associated symmetric matrix \( A \). The difference \( p - q \) is called the index of the form. By Sylvester's Law of Inertia, the signature is invariant under congruence transformations, meaning it is an intrinsic property of the form, independent of the coordinate system. A form is positive definite if \( q = 0 \) (all eigenvalues positive), negative definite if \( p = 0 \) (all eigenvalues negative), and indefinite if both \( p \) and \( q \) are non-zero. The determinant of \( A \) provides a quick check: for a \( 2 \times 2 \) form, \( \det(A) < 0 \) means indefinite, while \( \det(A) > 0 \) means definite (positive if \( A_{11} > 0 \), negative if \( A_{11} < 0 \)). In general, Sylvester's criterion says a form is positive definite if and only if all leading principal minors are positive. For example, the matrix \( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \) has signature \( (1,1) \) and represents a hyperbolic form.
Principal Axes and Diagonalization
To find the principal axes, we diagonalize the symmetric matrix \( A \) using an orthogonal transformation. Let \( \mathbf{y} = P^T \mathbf{x} \), where \( P \) is an orthogonal matrix whose columns are the normalized eigenvectors of \( A \). Then \( Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = \mathbf{y}^T (P^T A P) \mathbf{y} = \mathbf{y}^T \Lambda \mathbf{y} \), where \( \Lambda \) is a diagonal matrix containing the eigenvalues of \( A \). In the new coordinates \( \mathbf{y} \), the quadratic form becomes \( Q = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \dots + \lambda_n y_n^2 \). The axes \( y_1, y_2, \dots, y_n \) are the principal axes, and the eigenvalues \( \lambda_i \) determine the curvature along each axis. This transformation preserves lengths and angles because \( P \) is orthogonal, ensuring that the geometric shape of the level sets is unchanged, only rotated.
The derivations
Spectral Theorem for Self-Adjoint Operators
Let \( \hat{A} \) be a self-adjoint operator on a finite-dimensional complex inner-product space \( V \) with \( \dim V = n \), so that \( \hat{A}=\hat{A}^{\dagger} \). Then \( \hat{A} \) has real eigenvalues, its eigenvectors belonging to distinct eigenvalues are orthogonal, and \( V \) possesses an orthonormal basis \( \{|e_1\rangle,\dots,|e_n\rangle\} \) consisting entirely of eigenvectors of \( \hat{A} \). Equivalently, the unitary \( \hat{U} \) whose columns are those eigenvectors diagonalizes \( \hat{A} \): \( \hat{U}^{\dagger}\hat{A}\hat{U}=\hat{\Lambda}=\mathrm{diag}(\lambda_1,\dots,\lambda_n) \), and \( \hat{A}=\sum_{k}\lambda_k\,|e_k\rangle\langle e_k| \).
Determinant as the Unique Alternating Multilinear Form
On the space of \(n\times n\) matrices over a field \(K\), there is exactly one function \(D:(K^{n})^{n}\to K\) of the ordered columns \((\mathbf{a}_1,\dots,\mathbf{a}_n)\) that is (i) multilinear — linear in each column separately, (ii) alternating — \(D=0\) whenever two columns coincide, and (iii) normalized — \(D(\mathbf{e}_1,\dots,\mathbf{e}_n)=1\) on the standard basis. That unique function is the determinant \(\det\), it equals the Leibniz sum \(\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i}A_{\sigma(i)\,i}\), and from the characterization alone it obeys the product law \(\det(AB)=\det(A)\det(B)\).
A common pitfall is confusing the eigenvalues of \( A \) with the coefficients in the diagonalized form. The eigenvalues are the coefficients, but they are not necessarily the diagonal entries of the original matrix. Also, remember that the eigenvectors must be normalized and orthogonal to form the matrix \( P \); if they are not, the transformation is not orthogonal, and the geometric interpretation as a rotation is lost. C
Worked example
Problem. Consider the quadratic form \( Q(x,y) = 5x^2 + 6xy + 5y^2 \). Find its principal axes, signature, and the equation of the level set \( Q = 1 \) in the principal coordinate system.
- Write the matrix \( A \): \( A = \begin{pmatrix} 5 & 3 \\ 3 & 5 \end{pmatrix} \).
- Find the eigenvalues: \( \det(A - \lambda I) = (5-\lambda)^2 - 9 = 0 \), so \( \lambda = 2 \) or \( \lambda = 8 \).
- Find the eigenvectors: For \( \lambda = 2 \), \( \begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix} \mathbf{v} = 0 \), so \( \mathbf{v}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \). For \( \lambda = 8 \), \( \begin{pmatrix} -3 & 3 \\ 3 & -3 \end{pmatrix} \mathbf{v} = 0 \), so \( \mathbf{v}_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \).
- Construct \( P = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} \). The diagonalized form is \( Q = 2y_1^2 + 8y_2^2 \).
- The signature is \( (2,0) \), so the form is positive definite. The level set \( Q = 1 \) is \( 2y_1^2 + 8y_2^2 = 1 \), an ellipse with semi-axes \( 1/\sqrt{2} \) and \( 1/\sqrt{8} \).
Units check. If \( x, y \) are lengths, \( Q \) has units of length squared, consistent with the coefficients being dimensionless or having appropriate physical units.
Three quick tests
- What is the signature of the form \( Q = x^2 - y^2 \)?
Answer
The eigenvalues are \( 1 \) and \( -1 \), so the signature is \( (1,1) \). - If a quadratic form has signature \( (3,0) \), what is the shape of its level sets?
Answer
Ellipsoids, since the form is positive definite. - Why must the matrix \( A \) in \( Q = \mathbf{x}^T A \mathbf{x} \) be symmetric?
Answer
Because the coefficient of \( x_i x_j \) is \( A_{ij} + A_{ji} \), so we can always symmetrize \( A \) without changing \( Q \).
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-082 · D-079 · Threads: symmetry, waves, matter