the characteristic polynomial det(A-tI) is not the same as "the Jordan form". two matrices which are NOT similar, can have the same characteristic polynomial.

to determine if the Jordan forms are the same, you HAVE to look at the super-diagonal. it's the non-zero entries which matter.

you can think of a Jordan block as a sum: J = D+N, where D is a "diagonal part" and N is a "nilpotent part".

you can't tell if two matrices are similar just by looking at the characteristic polynomial. if they are similar, their characteristic polynomials

wil be the same, but NOT vice versa (it is a one way implication; you can have same diagonal parts, but different nilpotent parts).

you have to look at ker(A - tI), for each t. for A =

[a 0]

[0 a], ker(A -aI) is span({e1,e2}), dimension = 2. but for B =

[a 1]

[0 a], ker(B-aI) = span({e1}), which is a smaller kernel, dimension = 1.

in this case, we can show explicitly there is no invertible P for which PAP^-1 = B:

PAP^-1 = P(aI)P^-1 = a(PIP^-1) = a(PP^-1) = aI ≠ B, so A and B cannot be similar.

moral of the story: it is not enough to check that the characteristic polynomials are the same, you have to check the dimensions of the eigenspaces

(the dimension of the kernel ker(A-tI)), as well.

you can't just check the algebraic multiplicity of an eigenvalue, you have to check the geometric multiplicity, too. sometimes, they aren't the same.