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Suppose I have an eigenvalue $\lambda$, now I want to determine the number of Jordan blocks corresponding to that eigenvalue, as well as size of each block. I know that:
1) number of blocks is equal to the number of eigenvectors
2) there are $\dim \ker (A - \lambda I)^k - \dim \ker (A - \lambda I)^{k-1}$ blocks of size at least $k$
Why exactly are those statements true. Could someone provide me with a proof or at least some resoning behind this?

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To answer your first question, each block looks like

$$ \left( \begin{array}{ccccc} \lambda & 1 & & & \\ & \lambda & 1 & & \\ & & \ddots & \ddots & \\ & & & \lambda & 1 \\ & & & & \lambda \end{array} \right) $$

The columns of a matrix correspond to the basis vectors. You can see here that the first column gives an eigenvector, and the rest do not. The columns here tell you that

$$ \begin {align*} Ae_1 &= \lambda e_1 \\ Ae_2 &= e_1 + \lambda e_2 \\ &\vdots \\ Ae_n &= e_{n-1} + \lambda e_n \end {align*} $$

So each Jordan block gives you exactly 1 eigenvector.

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  • $\begingroup$ Great, that makes total sense, thank you. $\endgroup$ – G.Fil Sep 23 '17 at 22:01

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