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generalized eigenvector∗ CWoo† 2013-03-21 23:29:38 Let V be a vector space over a field k and T a linear transformation on V (a linear operator). A non-zero vector v ∈ V is said to be a generalized eigenvector of T (corresponding to λ) if there is a λ ∈ k and a positive integer m such that (T − λI)m (v) = 0, where I is the identity operator. In the equation above, it is easy to see that λ is an eigenvalue of T . Suppose that m is the least such integer satisfying the above equation. If m = 1, then λ is an eigenvalue of T . If m > 1, let w = (T − λI)m−1 (v). Then w 6= 0 (since v 6= 0) and (T − λI)(w) = 0, so λ is again an eigenvalue of T . Let v be a generalized eigenvector of T corresponding to the eigenvalue λ. We can form a sequence v, (T − λI)(v), (T − λI)2 (v), . . . , (T − λI)i (v), . . . , (T − λI)m (v) = 0, 0, . . . The set Cλ (v) of all non-zero terms in the sequence is called a cycle of generalized eigenvectors of T corresponding to λ. The cardinality m of Cλ (v) is its length. For any Cλ (v), write vλ = (T − λI)m−1 (v). Below are some properties of Cλ (v): • vλ is the only eigenvector of λ in Cλ (v), for otherwise vλ = 0. • Cλ (v) is linearly independent. Pm Proof. Let vi = (T − λI)i−1 (v), where i = 1, . . . , m. Let 0 = i=1 ri vi with ri ∈ k. Induct on i. If i = 1, then v1 = v 6= 0, so r1 = 0 and {v1 } is linearly independent. Suppose the property is P true when i = m − 1. m Apply T − λI to the equation, and we have 0 = i=1 ri (T − λI)(vi ) = Pm−1 i=1 ri vi+1 . Then r1 = · · · = rm−1 = 0 by induction. So 0 = rm vm = rm vλ and thus rm = 0 since vλ is an eigenvector and is non-zero. ∗ hGeneralizedEigenvectori created: h2013-03-21i by: hCWooi version: h39754i Privacy setting: h1i hDefinitioni h65F15i h65-00i h15A18i h15-00i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 • More generally, it can be shown that Cλ (v1 ) ∪ · · · ∪ Cλ (vk ) is linearly independent whenever {v1λ , . . . , vkλ } is. • Let E = span(Cλ (v)). Then E is a (m + 1)-dimensional subspace of the generalized eigenspace of T corresponding to λ. Furthermore, let T |E be the restriction of T to E, then [T |E ]Cλ (v) is a Jordan block, when Cλ (v) is ordered (as an ordered basis) by setting (T − λI)i (v) < (T − λI)j (v) whenever i > j. Indeed, for if we let wi = (T − λI)m+1−i (v) for i = 1, . . . m + 1, then λwi if i = 1, T (wi ) = (T − λI + λI)(T − λI)m+1−i (v) = wi−1 + λwi otherwise. so that [T |E ]Cλ (v) is the (m + 1) × (m + 1) matrix given by λ 0 0 .. . 0 0 1 λ 0 .. . 0 1 λ .. . ··· ··· ··· .. . 0 0 0 0 ··· ··· 0 0 0 .. . 1 λ • A cycle of generalized eigenvectors is called maximal if v ∈ / (T − λI)(V ). If V is finite dimensional, any cycle of generalized eigenvectors Cλ (v) can always be extended to a maximal cycle of generalized eigenvectors Cλ (w), meaning that Cλ (v) ⊆ Cλ (w). • In particular, any eigenvector v of T can be extended to a maximal cycle of generalized eigenvectors. Any two maximal cycles of generalized eigenvectors extending v span the same subspace of V . References [1] Friedberg, Insell, Spence. Linear Algebra. Prentice-Hall Inc., 1997. 2