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.


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.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much but unfortunately I am not able to understand your answer as I am just a student in my first semester of linear algebra. Would there be another way to explain it? $\endgroup$ – blank Jul 16 '17 at 17:31
  • $\begingroup$ This is indeed beyond first semester linear algebra (in France, this is in the curriculum of 3rd year students). I'm afraid there's no other other way. Maybe the structure theorem will be best understood if you take a look at the structure theorem for finitely generated abelian groups, which is the initial version of the general theorem. In any case, you may retain your criterion doesn't suffice to characterise similar matrices. $\endgroup$ – Bernard Jul 16 '17 at 17:49
  • $\begingroup$ I looked into both articles (although I understand the French words but not really the whole mathematical context) and it was quite overwhelming. Obviously I have a long way to go to finally grasp it. Thanks a lot anyway, you still helped me! And by the way, how do I show that two square non-diagonalizable matrices are similar then? $\endgroup$ – blank Jul 16 '17 at 18:23
  • $\begingroup$ If they have the same Jordan normal form, they're similar. Do you know about Jordan normal form? $\endgroup$ – Bernard Jul 16 '17 at 18:40
  • $\begingroup$ My lecturer told us of its existence but we won't deal with it. Therefore we are not allowed to use it and I don't really know how to solve this problem. $\endgroup$ – blank Jul 16 '17 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.