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I am not really clear on the generalized subspaced and eigenvectors. I know that generalized eigenvectors are those vectors v which satisfy $(A-λE)^k*v=0$ but I can't yet get my head around how to find them in practice.

Also another question: Is the generalized subspace supposed to have dim that equals to the algebraic multiplicity of the eigenvalue? So the generalized subspace of Jordan form would be something that has dim k? Or is it just the span of first vector of canonic basis?

Sorry for my english and lack of understanding :D

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    You find them by computing the nullspace of $(A - \lambda E)^k$, e.g. using row reduction. And yes, the space of generalized eigenvectors has dimension the algebraic multiplicity. – 2017-02-25
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    What is the generalized subspace of Jordan form J(x,k) for eigenvalue x then? – 2017-02-25
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    The entire codomain. – 2017-02-25

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