Finding generalized eigenvectors from a Jordan form I am trying to understand the relation between Jordan form, characteristic polynomial and minimal characteristic polynomial.
From: Problem5

Consider a matrix A, assume that A has characteristic polynomial
$$(s − \lambda _1)^5 (s − \lambda_2)^3$$
that it has four linearly independent eigenvectors and
has minimal polynomial $$(s − \lambda_1)^2(s − \lambda_2)^3.$$
Write down the Jordan form J of this matrix.

The answer available here says:

[...]
• The eigenvalue $\lambda_2$ will have only 1 block of 3 × 3, that contains only 1 eigenvector of A and two generalized eigenvectors.
• The eigenvalue $\lambda_1$ will have a block of 2 × 2, that contains only 1 eigenvector of A and one generalized eigenvector. [...]

How do I know how many eigenvectors are in each Jordan block? How many of  those are L.I.? How many of those are generalized? Is there a fast way to obtain the eigenvectors from a Jordan form matrix? Or should I just use the definition of eigenvector?
Also if you could suggest any online material on Jordan form properties and relation to eigenvectors that would be great.
 A: First I will answer your questions and then explain how to write J.
Each n$\times$n Jordan block represents one eigenvector and n − 1 generalized eigenvectors. All the eigenvectors and generalized eigenvectors are linearly independent. You cannot determine the eigenvectors from J unless J = A; you also need the matrix M from the similarity transformation J = M$^{-1}$AM.  Hence, yes, you can use the definition of an eigenvector applied to matrix A.  One online source on Jordan normal form properties and their relation to eigenvectors is the Wikipedia article "Generalized eigenvector," which also justifies my answers.
Here is how to write J.  The degree of the characteristic polynomial gives the size of J, 8$\times$8 in this case.  With the characteristic polynomial in the form you have it, the exponent of each monomial gives the number of times the corresponding eigenvalue appears along the diagonal of J; in this case, λ$_1$ appears five times and λ$_2,$ three times.  The number of linearly independent eigenvalues gives the number of Jordan blocks, four in this case.  With the minimal polynomial in the form you have it, the exponent of each monomial gives the size of the largest Jordan block for the corresponding eigenvalue.  In this case, λ$_2$ has a 3$\times$3 Jordan block, which takes care of all three λ$_2$'s along the diagonal of J.  The largest Jordan block for λ$_1$ is 2$\times$2, and we need three Jordan blocks for λ$_1.$  The only way to arrange that is to have two 2$\times$2 and one 1$\times$1 Jordan blocks. Thus, we can write J as shown in your answer document.
