# Generalised eigenspace contains the eigenspace?

I've been introduced to the definition of a generalised eigenspace for a linear operator $A$ of an n-dimensional vector space $V$ over an algebraically closed field $k$. If $\lambda_1, \lambda_2,...,\lambda_k$ are the eigenvalues of $A$ then the characteristic polynomial of $A$ is defined

$\chi_A(t) = det(tI - A) = \prod_i (t-\lambda_i)^{m_i}$

$m_i$ being the multiplicity of the eigenvalue $\lambda_i$. $\chi_A(A) = 0$.

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

$V(\lambda_k) = Im(q_k(A)) = (q_k(A)(v) \mid v \in V)$

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

I'm trying to show that the eigenspace $V_{\lambda_k}$ is contained in the generalised eigenspace $V(\lambda_k)$. 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 $v \in V_{\lambda_j}$, then $v = v_1 + ... v_k$ where $v_i \in V(\lambda_i)$.

$Av = \lambda_j v$

i.e.

$Av_1 + Av_2 + ... + Av_k = \lambda_j v_1 + \lambda_j v_2 + ... + \lambda_j v_k$

So one has $(A-\lambda_jI)v_i = 0 ~\forall i \leq k$

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?

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HallsofIvy
Homework Helper
The "generalized eigenspace" is the space of all generalized eigenvectors. It should be simple to show, from the definitions of "generalized eigenvector" and "eigenvector" that any eigenvector is a "generalized" eigenvector and so is in the space of generalized eigenvectors. What are those definitions?

The "generalized eigenspace" is the space of all generalized eigenvectors. It should be simple to show, from the definitions of "generalized eigenvector" and "eigenvector" that any eigenvector is a "generalized" eigenvector and so is in the space of generalized eigenvectors. What are those definitions?

Currently on mobile so I can't check this fully. If I recall a generalised eigenvector for $\lambda$ is a nonzero vector $v$ such that for some integer N

$(A-\lambda I)^N v = 0$

So obviously from they definition an eigenvector for lambda is inside the generalised eigenspace.

But how do I show it using the definition given in my original post? It would amount to showing the two statements of generalised eigenspace are the same. I can't seem to do that without having my original question answered. Showing that the eigenspace is inside the generalised eigenspace with my definition would mean showin either

1) An eigenvector is the image of some other vector under $q_k(A)$.

Or

2) It being an eigenvector automatically ensures all components of its direct sum this aren't part of the desired generalised eigenspace are zero.

I've tried and thought, I can't seem to do either.

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