Quadratic forms and conics
The energy stored in a circuit, the power dissipated and the ellipse traced by a Lissajous figure are all written the same way: as a quadratic form.
01Second-degree polynomials, written with a matrix
A quadratic form is a homogeneous polynomial of degree two: all its terms have degree two. In two variables, q(x, y) = a x² + b xy + c y². What makes it interesting is that it can be written with a symmetric matrix, splitting the cross term in half:
Writing it this way is not a notational whim: it turns a question about a polynomial —is it always positive? what curve does it describe?— into a question about a matrix, which we already know how to answer.
02The conics
Adding linear terms and a constant gives the general second-degree equation in two variables, which describes all the conics: the curves obtained by cutting a cone with a plane.
| Discriminant \( B^2 - 4AC \) | Conic | Standard form | Where it shows up |
|---|---|---|---|
| < 0, with A = C and B = 0 | Circle | \( x^{2} + y^{2} = r^{2} \) | Locus of constant magnitude; the \( |\Gamma| \) circle of the Smith chart |
| < 0 | Ellipse | \( \dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1 \) | Lissajous figure of two signals in quadrature; orbits |
| = 0 | Parabola | \( y^{2} = 4px \) | Parabolic antenna, projectile motion, power as a function of current |
| > 0 | Hyperbola | \( \dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1 \) | Constant-power curve \( v \cdot i = P \); time-difference location |
The discriminant B² − 4AC does not change when the axes are rotated: that is why it can be used to classify without solving anything. The limiting cases —two lines, a point, nothing— are called degenerate conics and appear when the plane passes through the vertex of the cone.
03The cross term is a rotation
If B ≠ 0, the conic is rotated with respect to the coordinate axes. You can always choose a pair of new axes, rotated by an angle θ, in which the cross term disappears:
Every symmetric matrix can be diagonalized with a rotation matrix. In the new axes the quadratic form has no cross terms and its coefficients are the eigenvalues: q = λ₁ x′² + λ₂ y′². That is all that is needed to classify it, and it is the same result that in mechanics gives the principal axes of inertia and in statistics, the principal components. It is developed in eigenvalues and eigenvectors.
04Classification by sign
A quadratic form is classified according to which signs it can take:
| Class | Condition on q(X) with X ≠ 0 | Eigenvalues | Leading principal minors |
|---|---|---|---|
| Positive definite | q > 0 always | All positive | All positive |
| Negative definite | q < 0 always | All negative | Alternate: −, +, −… |
| Semidefinite | q ≥ 0 (or ≤ 0), with some X ≠ 0 where it equals 0 | Some zero | Some zero |
| Indefinite | Takes values of both signs | Of both signs | Does not follow the patterns above |
Sylvester’s criterion lets you decide without computing eigenvalues: for 2 × 2, the form is positive definite if a > 0 and det(A) > 0.
q = 2x² + 2xy + 3y² has matrix [[2, 1], [1, 3]]: a = 2 > 0 and det = 6 − 1 = 5 > 0, so it is positive definite. It is never negative, and it equals zero only at the origin. Its eigenvalues are (5 ± √5)/2 ≈ 3.62 and 1.38, both positive, as they had to be.
05Seeing it
Change the coefficients and watch which curve appears, what the discriminant says and what happens to the eigenvalues and the rotation angle.
06Quadratic forms in engineering
- Energy. The energy stored in an inductor and a capacitor is E = ½L i² + ½C v²: a positive definite quadratic form in the state variables. Its being positive definite is what guarantees that the energy cannot be negative, and it is the basis of the Lyapunov stability criterion in control systems.
- Dissipated power. P = i² R, and in a network with several currents, P = IᵀRI. That it is positive definite is equivalent to saying that a passive network cannot deliver energy.
- Lissajous figures. Two sinusoidal signals of the same frequency trace on the oscilloscope in XY mode an ellipse whose tilt and flatness give the phase shift. It is a conic, and the cross term is precisely the phase shift. You can try it in the instrument simulator.
- Least squares. The error of a fit is a quadratic form in the parameters; the minimum exists and is unique precisely because it is positive definite.
In three variables, the same setup gives the quadrics: ellipsoids, hyperboloids of one and two sheets, elliptic and hyperbolic paraboloids —the “saddle”— and cones. The classification again comes out of the signs of the three eigenvalues.
07In the lab
For 5x² + 4xy + 5y² = 9: write the matrix, compute the discriminant and the eigenvalues, find the rotation angle and write the equation in the new axes. Check with the lab above.
With two generators of the same frequency in XY mode, vary the phase shift from 0° to 90° and record how the figure goes from a line to an ellipse and to a circle. Relate each case to the value of the cross term B.
Write the energy of an LC circuit as a quadratic form in (i, v), verify with Sylvester’s criterion that it is positive definite and discuss what it would mean if it were not.
08Common mistakes
- Putting the whole B off the diagonal instead of B/2: the matrix has to be symmetric and the product XᵀAX adds it twice.
- Classifying by looking only at A and C without considering the cross term.
- Confusing the discriminant with the determinant: B² − 4AC is −4·det(A), so the signs go the other way.
- Forgetting that rotation does not change the type of conic, only its orientation.
- Taking a form with det > 0 and a < 0 as positive definite: that is negative definite.
09Self-assessment
What matrix corresponds to q = 3x² − 4xy + y²?
[[3, −2], [−2, 1]]: the cross coefficient is split in halves.
Classify the conic 4x² + y² − 16 = 0.
B² − 4AC = 0 − 16 = −16 < 0 and A ≠ C: it is an ellipse, with semi-axes 2 in x and 4 in y.
Is q = x² + 4xy + y² positive definite?
No. The matrix is [[1, 2], [2, 1]], with det = 1 − 4 = −3 < 0: the eigenvalues are 3 and −1, of opposite signs, so the form is indefinite. For example q(1, −1) = −2.
What rotation angle eliminates the cross term in x² + 2xy + y²?
Since A = C, the angle is 45°. The conic is degenerate: (x + y)² = 0 is a double line.
Why is the energy of a passive circuit a positive definite form?
Because it is never negative and equals zero only when all currents and voltages are zero. If it could be negative, the circuit would deliver energy without a source.
10Further reading
- Charles H. Lehmann. Analytic Geometry. (Spanish edition, “Geometría analítica”), Limusa. The conics in detail: focal properties, rotation and translation of axes.
- Howard Anton. Elementary Linear Algebra. 5th ed. (Spanish edition, “Introducción al álgebra lineal”), Limusa Wiley, 2011. The chapter on quadratic forms, with the principal axes theorem and Sylvester’s criterion.
- Gilbert Strang. Linear Algebra and Its Applications. 4th ed. (Spanish edition, “Álgebra lineal y sus aplicaciones”), Cengage Learning, 2007. Relates positive definite matrices to minima, energy and stability.