Section 29.6 Theory
Here we will record some basis facts about generalized eignevectors. Note that these statements hold for every matrix, not just those with a single eigenvalue.
Theorem 29.6.1. Properties of generalized eigenspaces.
Suppose \(\lambda\) is an eigenvalue of the \(n\times n\) matrix \(A\text{,}\) with algebraic multiplicity \(m\text{.}\) Then the following hold.
- The generalized eigenspace of \(A\) corresponding to \(\lambda\) is a subspace of \(\R^n\) (or \(\C^n\text{,}\) as appropriate).
- Each generalized eigensubspace of \(A\) corresponding to \(\lambda\) is a subspace of the generalized eigenspace of \(A\) corresponding to \(\lambda\text{.}\) That is, \(E^k_\lambda(A) \subseteq G_\lambda(A)\) for all \(k \ge 1\text{.}\)
- Each generalized eigensubspace of \(A\) corresponding to \(\lambda\) is a subspace of the next. That is, \(E^k_\lambda(A) \subseteq E^{k+1}_\lambda(A)\) for all \(k \ge 1\text{.}\)
- The dimension of the generalized eigenspace of \(A\) corresponding to \(\lambda\) is equal to the algebraic multiplicity of \(\lambda\text{.}\) That is, \(\dim G_\lambda(A) = m\text{.}\)
- The generalized eigenspace of \(A\) corresponding to \(\lambda\) is equal to the \(\nth[m]\) generalized eigensubspace of \(A\) corresponding to \(\lambda\text{.}\) That is, \(G_\lambda(A) = E^m_\lambda(A)\text{.}\)
Proof of Statement 1.
You were asked to verify this using the Subspace Test in Discovery 29.4. Note that when you take two random vectors \(\uvec{u},\uvec{v}\) in \(G_\lambda(A)\) to begin to show that \(G_\lambda(A)\) is closed under addition, you cannot assume that they are both in the same null space. That is, you know that \(\uvec{u}\) is in the null space of \((\lambda I -A)^{k_1}\) for some positive integer \(k_1\text{,}\) and you know that \(\uvec{v}\) is in the null space of \((\lambda I -A)^{k_2}\) for some positive integer \(k_2\text{,}\) but you cannot assume that \(k_1\) and \(k_2\) are equal.
Proof of Statement 2.
This statement should be obvious from the definitions of \(G_\lambda(A)\) and \(E^k_\lambda(A)\text{.}\)
Proof of Statement 3.
We leave this as an exercise for you, the reader.
Proof of Statement 4.
We will state and prove this statement as part of Theorem 30.5.5 in Section 30.5. Alternatively, see Exercise 4 in §6.8 of [3] for the outline of a direct proof of this statement using the theory of abstract linear operators.
Proof of Statement 5.
This statement could be deduced from Statement 4 and the fact that \(\lambda I - A\text{,}\) when restricted to the subspace \(G_\lambda(A)\text{,}\) is a nilpotent operator. But we will not pursue that discussion here.