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A singular leading coefficient gives an infinite polynomial eigenvalue

Example

Take the grade-two matrix polynomial P(λ)=diag⁡(1,λ2), so n=m=2, det⁡P(λ)=λ2, and A2=diag⁡(0,1) is singular. The finite root λ=0 and the infinite root are different projective eigenvalues. Their algebraic and geometric multiplicities are computed directly below.

Facts & Assumptions

Given: A0=diag⁡(1,0), A1=0, and A2=diag⁡(0,1) for a regular grade-two polynomial.

[F1]

Regular matrix polynomials and homogeneous eigenpairs: Ph(1,0)=A2, and the algebraic multiplicity at infinity is the zero order at 0 of det⁡P2rev.

[F2]

First Frobenius companion pencil of a matrix polynomial: at infinity the homogeneous companion pencil is −Bc=diag⁡(A2,I2), with block eigenvectors (x,0)T for x∈ker⁡A2.

[F3]

The first Frobenius companion pencil preserves the finite and infinite polynomial spectrum: the finite and infinite eigenspaces correspond to those of the polynomial, and the infinite algebraic multiplicity is mn−deg⁡det⁡P.

Verification

technique · direct calculation
1.1givenF1F3algebra

The determinant is det⁡P(λ)=λ2≢0, so P is regular; its degree is 2, while mn=4. Thus the theorem's degree-defect formula gives infinite algebraic multiplicity 4−2=2. Also P2rev(μ)=diag⁡(μ2,1), whose determinant has a zero of order two at μ=0.

2.1givenF1F2step 1.1algebra

The leading matrix A2 has rank one and kernel span⁡{e1}, so the infinite eigenspace has geometric multiplicity one and is represented in the companion pencil by (e1,0)T. This differs from its algebraic multiplicity two.

3.1givenF1F2F3step 1.1algebra∎

At the finite root λ=0, P(0)=diag⁡(1,0) has kernel span⁡{e2}. The finite lift is (0,e2)T, which the companion theorem maps to a pencil eigenvector. Thus the finite root zero is not the infinite eigenvalue represented by (e1,0)T.

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