Matrix similarity equivalent to same characteristic polynomial and same geometric multiplicity I'm wondering whether the similarity of two square matrices is equivalent to them having the same characteristic polynomial and the same geometric multiplicity for each eigenvalue. It's obvious in case they are both diagonalizable but is it true when they are not? I can't seem to figure it out, any help would be nice.
 A: No, it's more complex than that. One can find  matrices with the same (single) eigenvalue, the same characteristic and minimal polynomials, yet non-similar. 
There exists a complete set of similarity invariants which characterises similar matrices. It is based on the structure theorem for finitely generated modules over P.I.D.s :

For any finitely generated module $M$ over a P.I.D. $R$, there is a sequence of elements $(d_1,d_2,\dots , d_n)$ such that
   $$M\simeq R/(d_1)\times R/(d_2)\times\dots\times R/(d_n) \quad\text{and}\quad d_1\mid d_2\mid \dots\mid d_n $$
  Furthermore the $d_i$s are unique but for a unit factor. They're called the invariant factors of the module $M$.

Now let $A$ be an $n\times n$ matrix over the field $K$. The vector space $K^n$ can be seen as a $K[X]$-module in the following way: for any vector $u\in K^n$, one sets
$$X\cdot u\overset{\text{def}}{=} Au. $$
One easily checks this turns $K^n$ into a finitely generated module over the P.I.D. $K[X]$. The invariant factors of this module are the similarity invariants of the matrix $A$.
You can have more details on Wikipedia.fr if you can read French. Unfortunately, there doesn't seem be a similar (;o)) Wikipedia article in English.
