Generalized eigenvectors and eigenspaces

Definition: Let \(\alpha\in End(V)\) and \(\lambda\in spec(\alpha)\). A non-zero vector \(v\) is called a generalized eigenvector vector of \(\alpha\) associated with \(\lambda\) if \((\alpha-\lambda I)^{k}(v)=0\) and \((\alpha-\lambda I)^{k-1}(v)\ne 0\) for some \(k\ge 1\) where \(k\) is called the degree of nilpotence for \(v\).

Let \(\lambda\in spec(\alpha)\). Then, \(M_{\lambda}=\bigcup\limits_{m=1} ker(\lambda I-\alpha)^m\) is what we call it the generalized eigenspace corresponding to \(\lambda\). Clearly, \(M_{\lambda}\) is the union of the zero vector and the set of all generalized eigenvectors of \(\alpha\) associated with \(\lambda\)

Fact: \(M_{\lambda}\) is a subspace and \(\alpha-\)invariant and if \(v\) is a generalized vector of index \(k\) then \(\{v,(\alpha-\lambda I)v,\cdots, (\alpha-\lambda I)^{k-1}v\}\) is linearly independent.

Proof: Let \(a\in \mathbb{F}\) and let \(v,w\in V\) be generalized eigenvectors of \(\alpha\) associated with \(\lambda\) of degrees \(k\) and \(h\) respectively. Then,

\((\alpha-\lambda I)^k(v)=0\) and \((\alpha-\lambda I)^h(w)=0\)

\(\implies v\in ker (\alpha-\lambda I)^k\) and \(w\in ker(\alpha-\lambda I)^h\)

\(\implies v\in ker (\alpha-\lambda I)^{k+h}\) and \(w\in ker(\alpha-\lambda I)^{k+h}\) because \((\alpha-\lambda I)^{k+h}(v)=0\) and \((\alpha-\lambda I)^{k+h}(w)=0\)

\(\implies v+w \in ker(\alpha-\lambda)^{k+h}\)

And, \((\alpha-\lambda I)^{k+h}(av)=a.(\alpha-\lambda I)^{k+h}(v)=0\).

This implies that \(M_{\lambda}\) is a subspace of \(V\).

Invariance: If \(\beta\in End(V)\) commutes with \(\alpha\) and if \(v\) is a generalized eigenvector of \(\alpha\) associated with \(\lambda\) such that \(v\in ker(\alpha-\lambda I)^k\) then,

\((\alpha-\lambda I)^k\beta(v)=\beta(\alpha-\lambda I)^k(v)=0_V\)

\(\implies \beta(v)\) is also a generalized eigenvector of \(\alpha\) associated with \(\lambda\)

Linearly Independence: If \(v\) is a generalized vector of \(\alpha\) associated with \(\lambda\) then \((\alpha-\lambda I)^k(v)=0\) and \((\alpha-\lambda I)^{k-1}(v)\ne 0\). Now we assume that,

\(c_0v+c_1(\alpha-\lambda I)(v)+\cdots+c_{k-1}(\alpha-\lambda I)^{k-1}(v)=0\).

We need to show that \(c_i's\) are zero for \(0\le i\le k-1\).

Applying \((\alpha-\lambda)^{k-1}\) we get,

\((\alpha-\lambda)^{k-1}(c_0v+c_1(\alpha-\lambda I)(v)+\cdots+c_{k-1}(\alpha-\lambda I)^{k-1}(v))=0\)

\(\implies c_0(\alpha-\lambda I)^{k-1}(v)=0\). Since \((\alpha-\lambda I)^{k-1}(v)\ne 0\) so we get \(c_0=0\).

Similarly applying \((\alpha-\lambda I)^{k-2},(\alpha-\lambda I)^{k-3},\) and so on, we have

\(c_i=0\) for \(1\le i\le k-1\).

Hence, \(\{v,(\alpha-\lambda I)v,\cdots, (\alpha-\lambda I)^{k-1}v\}\) is linearly independent.

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