Recipe

Find the principal axes and standard form of a conic or quadric

Use when: you have a two- or three-variable quadratic expression with a nonzero cross term (xy, xz, or yz) and need its axes or its cross-term-free standard form

Goal: the orthonormal eigenbasis that removes the cross term, plus the standard form in that basis

Taught in Conics, quadrics, and the principal-axis theorem

1. 1

   Put each squared-term coefficient on the matching diagonal entry. Split every cross-term coefficient in half, placing one half on each of the two matching off-diagonal entries.

   Example

   For 5x^2-4xy+8y^2, the diagonal entries are 5 and 8, the squared-term coefficients. The xy coefficient -4 splits in half: A=\begin{bmatrix}5&-2\\-2&8\end{bmatrix}

   Why: x^{\mathsf T}Ax only reproduces the original form when each off-diagonal entry is half the cross-term coefficient, since symmetry counts that entry twice.

2. 2

   Solve the characteristic equation \det(A-\lambda I)=0; its roots are the coefficients the standard form will use.

   Example

   \det(A-\lambda I)=(5-\lambda)(8-\lambda)-4=\lambda^2-13\lambda+36=(\lambda-4)(\lambda-9)=0, so \lambda=4 or \lambda=9.

   ⚡ Shortcut

   For a 2\times2 matrix, skip straight to \lambda^2-\operatorname{tr}(A)\lambda+\det(A)=0: here \operatorname{tr}(A)=13 and \det(A)=36, the identical equation without expanding (5-\lambda)(8-\lambda) by hand.

   Why: the principal-axis theorem above shows these eigenvalues become exactly the coefficients multiplying y_1^2 and y_2^2 once the cross term is gone.

3. 3

   Solve (A-\lambda I)v=0 for each root, then divide each eigenvector by its own length so every column has length 1.

   Example

   For \lambda=4: (A-4I)v=0 gives v=(2,1), length \sqrt5, so the unit eigenvector is (2/\sqrt5,\,1/\sqrt5). For \lambda=9: v=(-1,2), length \sqrt5, unit eigenvector (-1/\sqrt5,\,2/\sqrt5).

   • eigenvalues are distinct — the eigenvectors are automatically orthogonal, so only normalizing is needed.
   • an eigenvalue is repeated — its eigenspace has dimension 2 or more, so run Gram-Schmidt inside that eigenspace to pick an orthonormal basis for it.

   Why: Q's columns must be unit length and mutually orthogonal, since that is exactly what makes x=Qy a pure rotation with no stretching or shearing.

4. 4

   Put the unit eigenvectors into Q's columns, in the same order as the eigenvalues, then write the form as \sum_i\lambda_iy_i^2 using those eigenvalues.

   Example

   Q=\begin{bmatrix}2/\sqrt5&-1/\sqrt5\\1/\sqrt5&2/\sqrt5\end{bmatrix}

   Setting 5x^2-4xy+8y^2=36 becomes 4y_1^2+9y_2^2=36, the axes pointing along Q's two columns.

   Why: substituting x=Qy into x^{\mathsf T}Ax collapses it to y^{\mathsf T}(Q^{\mathsf T}AQ)y=y^{\mathsf T}\Lambda y exactly, as this section's derivation shows.

5. 5

   Compare the signs of the eigenvalues to classify the curve or surface, now that the cross term is gone.

   Example

   4y_1^2+9y_2^2=36 has both eigenvalues positive, so dividing through gives y_1^2/9+y_2^2/4=1: an ellipse with semi-axes 3 and 2.

   • all eigenvalues share the same sign — an ellipse (in two variables) or ellipsoid (in three).
   • the eigenvalues have mixed signs — a hyperbola (in two variables) or a hyperboloid (in three).

   Why: once no cross term remains, each term's sign alone decides whether increasing that coordinate adds to or subtracts from the total, which is exactly what separates these shapes.