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Every regular complex pencil has a homogeneous eigenvalue pair

Statement

Every regular finite square pencil $(A, B)$ over $R$ or $C$ has at least one homogeneous generalized eigenvalue pair $[α : β]$ over $C$ .

Facts & Assumptions

Given: $n \geq 1$ and a regular pencil $A - λ B$ with $A, B \in F^{n \times n}$ , $F \in {R, C}$ .

[F1]

Regular matrix pencils and homogeneous generalized eigenvalues: regularity means $f (λ) = \det (A - λ B)$ is not the zero polynomial, and a homogeneous eigenvalue pair is a nonzero projective zero of $p (α, β)$ .

[F2]

The pencil determinant is homogeneous and its zero condition is scale invariant: $p (α, β) = β^{n} f (α / β)$ when $β \neq 0$ .

[F3]

Fundamental theorem of algebra: every nonconstant complex polynomial has a complex root: every nonconstant complex polynomial has a complex root.

[F4]

The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring: the determinant is multilinear in its columns.

Proof

1.1givenF1F4algebra

By column multilinearity [F4], expanding the $n$ columns of $A - λ B$ expresses $f (λ) = \det (A - λ B)$ as a polynomial of degree at most $n$ . Regularity [F1] makes it nonzero.

2.1step 1.1F2F3constructassume-case degree

If $\deg f = n$ , [F3] gives $λ_{0} \in C$ with $f (λ_{0}) = 0$ . Set $[α : β] = [λ_{0} : 1]$ ; [F2] gives $p (α, β) = 0$ .

2.2step 1.1F1F4algebraconstructassume-case lower

If $\deg f < n$ , its coefficient of $λ^{n}$ is zero. In the same column-multilinear expansion, that coefficient is $(- 1)^{n} \det B$ , obtained by taking the $- λ B$ column in every position. Hence $\det B = 0$ . Setting $[α : β] = [1 : 0]$ gives $p (1, 0) = \det (- B) = (- 1)^{n} \det B = 0$ , so this is an infinite homogeneous eigenvalue pair.

3.1step 2.1step 2.2cases-exhaustive∎

The two degree cases exhaust the nonzero polynomial $f$ and each produces a nonzero projective pair with $p (α, β) = 0$ .

Depends on

Used by

Every regular real or complex pencil has a generalized Schur form Theorem

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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