Recipe

Find an ellipsoid's semi-axes and read its shape from a positive definite matrix

Use when: you have a positive definite matrix A and need the semi-axis lengths, axis directions, or overall shape of the level set \{x^{\mathsf T}Ax=1\}

Goal: every semi-axis length, its matching direction, and a shape comparison based on how spread out the eigenvalues are

Taught in Ellipsoids and the shape of a positive definite matrix

1. 1

   Solve \det(A-\lambda I)=0 if A isn't diagonal; a diagonal matrix's eigenvalues are already sitting on the diagonal.

   Example

   For A=\begin{bmatrix}5&-2\\-2&8\end{bmatrix}, \det(A-\lambda I)=\lambda^2-13\lambda+36=(\lambda-4)(\lambda-9)=0, so \lambda=4 or \lambda=9.

   • A is diagonal — read the eigenvalues straight off the diagonal entries, skipping the characteristic equation entirely.
   • A isn't diagonal — solve the characteristic equation, exactly as in the principal-axis recipe.

   Why: the level set reduces to \sum_i\lambda_iy_i^2=1 in principal-axis coordinates, so every semi-axis length below is built from these numbers.

2. 2

   Take the reciprocal square root of each eigenvalue: 1/\sqrt{\lambda_i}.

   Example

   \lambda=4 gives 1/\sqrt4=0.5; \lambda=9 gives 1/\sqrt9\approx0.333.

   Why: along the i-th principal axis alone the equation reduces to \lambda_iy_i^2=1, forcing y_i=\pm1/\sqrt{\lambda_i}.

3. 3

   Find the unit eigenvector for each eigenvalue; that direction is where its semi-axis length actually points.

   Example

   \lambda=4's unit eigenvector is (2/\sqrt5,\,1/\sqrt5), so the ellipse reaches 0.5 along that direction; \lambda=9's is (-1/\sqrt5,\,2/\sqrt5), reaching 0.333 along it.

   • the eigenvalues are distinct — each direction is unique up to a sign flip.
   • an eigenvalue is repeated — the ellipsoid is a perfect sphere inside that eigenspace, so any orthonormal basis of it works as axis directions.

   Why: the semi-axis length alone only says how far the level set reaches, not which way, so it needs the matching eigenvector to mean anything geometrically.

4. 4

   Compute the ratio \lambda_{\max}/\lambda_{\min} across all the eigenvalues found in Step 1.

   Example

   Here \lambda_{\max}/\lambda_{\min}=9/4=2.25, a moderate spread: noticeably stretched, but far from extreme.

   • the ratio sits close to 1 — the eigenvalues are all similar sizes, so the ellipsoid is nearly a sphere.
   • the ratio is far from 1 — the ellipsoid is stretched or flattened along the small-eigenvalue direction.

   Why: this ratio is the same condition-number comparison chapter 10 used for singular values, and it measures exactly how uneven the axis lengths are.