Generalized Eigenvalue Problem
Solve A v=lambda B v for symmetric 2x2 A and positive diagonal B.
Free theory articleSolve A v=lambda B v for symmetric 2x2 A and positive diagonal B.
Concept & interpretation
The symmetric generalized problem is Av = λBv, where B = diag(b1,b2) must be positive definite. Set C = B−1/2AB−1/2. Solve Cu = λu with Euclidean unit vectors, then v = B−1/2u satisfies vTBv = 1. The app sorts eigenvalues and makes the first nonzero vector component positive.
Reproduce and compare both eigenpairs
Open Generalized Eigenvalue Problem. The plot uses one shared auto-fit scale for the green v and blue Av; pixel lengths across settings do not establish vector magnitudes.
- Use Reset, then set A11 to 2, A12 to 1 and A22 to 2.
- Set b1 and b2 to 1. Select Lower in Eigenpair: λ = 1 and v ≈ [0.707107,−0.707107]. Check norm 1 and near-zero residual.
- Select Upper: λ = 3 and v ≈ [0.707107,0.707107].
- Set b1 = b2 = 2. The eigenvalues become 0.5 and 1.5, while the selected upper vector is [0.5,0.5] and its B-norm remains 1.
- Set b1 to 0. The app reports that the metric is not positive definite and hides eigenpairs. Reset restores the original valid diagonal example.