Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism. If \(v\) is a generalised eigenvector of \(f\) of rank \(m\) with respect to the eigenvalue \(\lambda,\) then \((f-\lambda \mathrm{Id}_V)^{m-1}v\) is an eigenvector of \(f\) with respect to \(\lambda.\)

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism. If \(v\) is a generalised eigenvector of \(f\) of rank \(m>1,\) it also must be an eigenvector of \(f.\)

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism. If \(v\) is a generalised eigenvector of \(f\) of rank \(m>1\) with respect to eigenvalue \(\lambda,\) then there exists \(0_V\ne v\in V\) such that \(f(v)=\lambda v.\)

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism. If \(v\) is a generalised eigenvector of \(f\) of rank \(m>1,\) then there exists a generalised eigenvector of rank \(m-1\) with respect to the same eigenvalue.

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism with eigenvalues \(\lambda_1,\ldots,\lambda_k,\) then \[\operatorname{Eig}_f(\lambda_1)\oplus\cdots\oplus\operatorname{Eig}_f(\lambda_k) = \mathcal E_f(\lambda_1)\oplus\cdots\oplus\mathcal E(\lambda_k)\]

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be a diagonalisable endomorphism with eigenvalues \(\lambda_1,\ldots,\lambda_k,\) then \[\operatorname{Eig}_f(\lambda_1)\oplus\cdots\oplus\operatorname{Eig}_f(\lambda_k) = \mathcal E_f(\lambda_1)\oplus\cdots\oplus\mathcal E(\lambda_k)\]

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism with eigenvalue \(\lambda.\) It always holds that \[\dim(\operatorname{Eig}_f(\lambda))<\dim(\mathcal E_f(\lambda))\].

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism. If \(U\subset V\) is stable under \(f,\) then \(U\) must be a generalised eigenspace of \(f.\)

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Let \(V\) be a finite dimensional complex vector space and let \(f:V\to V\) be an endomorphism such that \(f\circ f = f.\) Then \(f\) does not admit a generalised eigenvector of rank \(2.\)

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Let \(V\) be a finite dimensional complex vector space and let \(f,g:V\to V\) be endomorphisms. If \(f\) and \(g\) share the same generalised eigenspaces, they must be similar.

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