# Are there infinitely many equivalence classes of similar matrices? [closed]

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.

-

## closed as too localized by Mariano Suárez-Alvarez♦, Harald Hanche-Olsen, Robin Chapman, Harry Gindi, Qiaochu YuanApr 29 '10 at 23:25

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The determinant of two similar matrices is the same. Since there are infinitely many different possible determinats (when the field is infinite!), then there are infinitely many similarity classes. – Mariano Suárez-Alvarez Apr 29 '10 at 19:54
You should probably look in the FAQ at mathoverflow.net/faq#whatnot for a list of sites where this question, and other similar (!) ones will make a much better fit. This site, as explained in the FAQ, has a slightly different focus. – Mariano Suárez-Alvarez Apr 29 '10 at 19:55

## 1 Answer

[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.

-