Are there infinitely many equivalence classes of similar matrices? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-20T12:03:37Z http://mathoverflow.net/feeds/question/23029 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23029/are-there-infinitely-many-equivalence-classes-of-similar-matrices Are there infinitely many equivalence classes of similar matrices? To be cont'd 2010-04-29T19:51:34Z 2010-06-10T20:08:17Z <p>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.</p> <p>Thanks in advance for any comment.</p> http://mathoverflow.net/questions/23029/are-there-infinitely-many-equivalence-classes-of-similar-matrices/23042#23042 Answer by Pete L. Clark for Are there infinitely many equivalence classes of similar matrices? Pete L. Clark 2010-04-29T21:14:31Z 2010-04-29T21:14:31Z <p>[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.] </p> <p>For a positive integer $n$, consider the ring $M_n(k)$ of $n \times n$ matrices with $k$-coefficients for $n \geq 1$. </p> <p>If $k$ is finite, then $M_n(k)$ is finite, so obviously there are only finitely many similarity classes.</p> <p>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.</p> <p>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 $k$?</p> <p>When $k$ is infinite, it follows easily from the above that <code>$S(n,k) = \# k$</code>.</p> <p>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.</p>