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A singular B can give a regular pencil with an infinite generalized eigenvalue

Example

Take $A = I_{2}$ and $B = diag (1, 0)$ . The matrix $B$ is singular, but the pencil is regular and has one finite eigenvalue and one infinite eigenvalue.

Facts & Assumptions

Given: The pair $(A, B)$ and the homogeneous eigenvalue convention of Regular matrix pencils and homogeneous generalized eigenvalues.

[F1]

Regular matrix pencils and homogeneous generalized eigenvalues: regularity means $\det (A - λ B)$ is not identically zero, and homogeneous eigenpairs solve $\det (β A - α B) = 0$ with $(α, β) \neq (0, 0)$ .

Verification

1.1givenF1algebra

Direct calculation gives $\det (A - λ B) = 1 - λ$ , a nonzero polynomial, so the pencil is regular although $\det B = 0$ .

2.1F1step 1.1algebra∎

The homogeneous determinant is $\det (β A - α B) = \det (\begin{array}{cc} β - α & 0 \\ 0 & β \end{array}) = β (β - α) .$ Its projective roots are $[1 : 1]$ , the finite eigenvalue $1$ , and $[1 : 0]$ , the infinite eigenvalue.

Depends on

Regular matrix pencils and homogeneous generalized eigenvalues

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Peter Arbenz, Lecture Notes on Solving Large Scale Eigenvalue Problems, §§12.8–12.9 (standard reference, not scraped)
LAPACK Users' Guide, Generalized Nonsymmetric Eigenproblems (standard reference, not scraped)