It is easy to show that similarity in matrices is an equivalence relation (two matrices A and B of same size being similar if there exists a matrix P such that B = PAP^(-1) )
Moreover, given a matrix, its equivalence class can be finite. E.g. The equivalence of nxn matrices containing the identity matrix I is singleton (i.e. it contains only the identity matrix itself).
But I do not know how many equivalence classes there are for matrices of a given size.


Thanks in advance for any comment.



Are there infinitely many equivalence classes of similar matrices?
[This is an easy question, but it doesn't feel like a homework question, so I will answer it. I have made the post community wiki to protect myself from unwanted votes, both upwards and downwards.] 


For a positive integer $n$, consider the ring $M\_n(k)$ of $n \times n$ matrices with $k$-coefficients for $n \geq 1$. 


If $k$ is finite, then $M\_n(k)$ is finite, so obviously there are only finitely many similarity classes.


If $k$ is infinite, then since the determinant map $M\_n(k) \rightarrow k$ is surjective and the determinant is a similarity invariant, there are infinitely many similarity classes.


One may ask a more precise question: what is the cardinality $S(n,k)$ of the set of 
similarity classes of $n \times n$ matrices with coefficients in $k$?


When $k$ is infinite, it follows easily from the above that $S(n,k) = \# k$.


On the other hand, when $k \cong \mathbb{F}\_q$ is finite, it is a nice linear algebra exercise to give an explicit formula for $S(n,k)$ in terms of $q$ and $n$. It might (or might not) be appropriate to discuss how to derive such a formula here.


