## 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.

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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 2010 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 2010 at 19:55

## closed as too localized by Mariano Suárez-Alvarez, Harald Hanche-Olsen, Robin Chapman, Harry Gindi, Qiaochu YuanApr 29 2010 at 23:25

[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.

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