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The pencil determinant is homogeneous and its zero condition is scale invariant

Statement

For $A, B \in F^{n \times n}$ , $n \geq 1$ , let $p (α, β) = \det (β A - α B)$ . For every $c \in C$ , $p (c α, c β) = c^{n} p (α, β) .$ Consequently the zero condition is independent of the nonzero representative of a homogeneous pair. If $β \neq 0$ , then $p (α, β) = β^{n} \det (A - (α / β) B) .$

Facts & Assumptions

Given: $n \geq 1$ , $A, B \in F^{n \times n}$ , and $(α, β) \in C^{2}$ .

[F1]

Regular matrix pencils and homogeneous generalized eigenvalues: $p (α, β)$ is $\det (β A - α B)$ , and homogeneous pairs are identified by nonzero complex scaling.

[F2]

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

Proof

1.1givenF1F2algebra

Scaling $(α, β)$ by $c$ scales every column of $β A - α B$ by $c$ . Applying [F2] to its $n$ columns gives $\det (c (β A - α B)) = c^{n} \det (β A - α B)$ , which is the first identity by [F1].

2.1step 1.1F1F2algebra∎

If $c \neq 0$ , then $c^{n} \neq 0$ , so $p (c α, c β) = 0$ exactly when $p (α, β) = 0$ . If $β \neq 0$ , factor $β A - α B = β (A - (α / β) B)$ and apply the same column multilinearity to obtain the finite-eigenvalue identity.

Depends on

Used by

Every regular complex pencil has a homogeneous eigenvalue pair Lemma

Dependency tree · two levels

7 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)