Let $M_n(K)$ be the set of all $n\times n$ matrices over a field $K$. If $\mathcal{R}$ is the equivalence relation defined by matrix similarity, what does the quotient $M_n(K)/\mathcal{R}$ looks like? Is there something that characterizes it in terms of cardinality? Is there a way to extend matrix operations to equivalence classes, making it an algebraic quotient structure of $M_n(K)$? Is there a good interpretation of $M_n(K)/\mathcal{R}$ in terms of similarity invariants (e.g., matrix determinants)?

Today, on my linear algebra class, we were introduced the concept of determinants. Our teacher used geometric isometry invariants (such as area and volume) to introduce the motivation for matrix determinants. I then asked myself what would be the generalized interpretation of the determinant as some type of ''invariant''. Some Google search led me to matrix similarity, and my algebraic intuition says that this information would be codified in the quotient structure - once we ''kill' the big structure by the invariants, we would find exactly what is not varying.

Well, though I never though about verifying it for any pathological case, I believe that the map $\det: M_n(K) \to K$ is surjective. That would mean that the quotient $M_n(K)/\mathcal{R}$ is at least the same cardinality as $K$ (because similar matrices have the same determinant, but I'm not sure about the converse. If it is true, matrix with the same determinant are similar, then cardinality equality would follow).

I'm not sure how the operations can be extended. Maybe the quotient can be seen as a vector space over $K$ as well: scalar multiplication is surely well defined, but I'm not sure about the sum. If I restrict myself to some subset of $M_n(K)$, like the general linear group or the special linear group, could I get something more?

And the interpretation or mathematical application is exactly what I long for.

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    $\begingroup$ This can be generalized to any algebra or ring. Given a (not necessarily commutative) ring $R$, for elements $a, b \in R$, define $a \sim b$ iff $x^{-1}ax = b$ (or $ax = xb$) , for some unit $x$ of $R$. The equivalence classes $R / \sim$ can also be seen as the orbits of the action of $U$ (the group of units of $R$) on $R$ by conjugation. Of course, this is trivial for a commutative ring. $\endgroup$ – M. Vinay Jun 1 '16 at 3:22

Let $\mathcal C$ be the set of $n\times n$ matrices over $K$ which are in rational canonical form. Since every element of $M_n(K)$ is similar to a unique matrix in rational canonical form, you can identify $M_n(K)/\mathcal R$ with $\mathcal C$. In particular, the cardinality of $M_n(K)/\mathcal R$ is the number of $n\times n$ matrices over $K$ in rational canonical form, i.e. the number of ways to construct a matrix $$ \begin{bmatrix} 0&0&\dots&\dots&\dots&-b_0\\ 1 &0&\dots&\dots&\dots&-b_1\\ 0&1&\dots&\dots&\dots&-b_2\\ 0&0&\ddots& &&\vdots\\ \vdots&\vdots&&\ddots&&\vdots\\ 0&0&\dots&\dots&1&-b_{n-1} \end{bmatrix}. $$ There are $n$ choices for $b_0,\dots,b_{n-1}$, and so $$ |M_n(K)/\mathcal R|=|K^n|. $$ As for your reasoning about the determinant: it is true that $\det:M_n(K)\to K$ is a surjective homomorphism. Given $x\in K$, the matrix $$ \begin{bmatrix} x&0&\dots&0\\ 0&1&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&1 \end{bmatrix} $$ has determinant $x$. Since similar matrices have the same determinant, you are right that $|M_n(K)/\mathcal R|\geq |K|$. However, the converse is false: matrices with the same determinant might not be similar. For example, $$ SO(2)=\left\{\begin{bmatrix}\cos\theta &-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}:\theta\in [0,2\pi)\right\} $$ consists of rotation matrices with determinant $1$, but no two distinct matrices in $SO(2)$ are similar (one way to see prove this is to note that the eigenvalues of a rotation matrix by $\theta$ are $e^{\pm i\theta}$).

  • $\begingroup$ Thanks for pointing it! Sadly, I guess those matrices in the canonical form don't have a natural way to be composed into a algebraic structure (well, just looking at it, matrix addition and multiplication won't work). Though it leaves the question that motivated mine still open (determinants are invariants of exactly what?), it gives an answer to most of my questions. I'll look now to what happens when we kill $M_n(K)$ by a equivalence relation defined on $A \sim B \iff \det(A) = \det(B)$. If I fail to find something, it may be material for another question. $\endgroup$ – Henrique Augusto Souza Jun 1 '16 at 2:39
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    $\begingroup$ @HenriqueAugustoSouza Observe that "similarity" is a refinement of "determinantal equality" (as equivalence relations). $\endgroup$ – M. Vinay Jun 1 '16 at 2:43
  • $\begingroup$ @M.Vinay that is certainly true, once that similarity implies determinant equality, but what I would also like to know is maybe how ``determinant equality" is a rudeness of similarity (if such term even makes sense). Sufficient and necessary conditions for determinant equality, from a general, abstract point of view. Maybe I won't be able to tell much about this such quotient to get that result (I believe that the other way around is more probable), but it can tell me more about determinants specifically (in contrast with other similarity invariants) $\endgroup$ – Henrique Augusto Souza Jun 1 '16 at 2:49
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    $\begingroup$ @HenriqueAugustoSouza Agreed, I'm not aware of any natural algebraic structure you could put on $\mathcal C$. As for the other question: determinants capture how matrices "scale volume." That is, if you apply a matrix $A\in M_n(\mathbb R)$ to each of the standard basis vectors in $\mathbb R^n$ (forming a unit $n$-cube), the signed volume of the $n$-parallelepiped formed from the output vectors is $\det(A)$. In my view this is the "right" way to view the determinant, since it makes certain properties clearer (it's similarity-independent, linear dependence of rows implies zero determinant, etc.) $\endgroup$ – Michael M Jun 1 '16 at 2:53

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