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.
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, \(G_\lambda(A) = E^m_\lambda(A)\text{.}\)
You were asked to verify this using the Subspace Test (Procedure 17.3.1) 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.
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.
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.
