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[itex] \chi_A(t) = det(tI - A) = \prod_i (t-\lambda_i)^{m_i} [/itex]

[itex] m_i [/itex] being the multiplicity of the eigenvalue [itex] \lambda_i [/itex]. [itex] \chi_A(A) = 0 [/itex].

Define [itex] q_k(t) = \prod_{i \neq k}(t-\lambda_i)^{m_i} = \chi_A(t)/(t-\lambda_k)^{m_k}[/itex]. Then we define the generalised eigenspace [itex] V(\lambda_k) [/itex] as

[itex] V(\lambda_k) = Im(q_k(A)) = (q_k(A)(v) \mid v \in V) [/itex]

One can show that [itex] V [/itex] is the direct sum of these generalised eigenspaces. By construction [itex] V(\lambda_k) [/itex] is [itex] A [/itex] invariant and [itex] (A-\lambda_k I)^{m_k}(v) = 0 ~\forall v \in V(\lambda_k) [/itex]

I'm trying to show that the eigenspace [itex] V_{\lambda_k} [/itex] is contained in the generalised eigenspace [itex] V(\lambda_k) [/itex]. The worst bit is my text says its straightforward, and to top it off I've actually done this before but forgotten how.

I'm honestly a bit lost as to how to do this. Take [itex] v \in V_{\lambda_j} [/itex], then [itex] v = v_1 + ... v_k [/itex] where [itex] v_i \in V(\lambda_i) [/itex].

[itex] Av = \lambda_j v [/itex]

i.e.

[itex] Av_1 + Av_2 + ... + Av_k = \lambda_j v_1 + \lambda_j v_2 + ... + \lambda_j v_k [/itex]

So one has [itex] (A-\lambda_jI)v_i = 0 ~\forall i \leq k [/itex]

But can anyone help point me in the right direction as to how I would go about proving this inclusion, using

*this*definition of generalised eigenspace?