# Matrices with eigenvalues conjugacy classes in Special Linear Group

By special linear group, I mean matrices with determinant equal to one.

I know a conjugacy class has to be equal to the identity as the identity commutes with other elements. I also know that similar matrices have the same determinant trace and eigenvalues.

How do I find the number of conjugacy classes? and the elements that represent each class? I am familiar with spectral theorem and orbits and stabilizers but don't see how they could be of use.

• Can you tell us anything about the coefficients of the matrices? Are they complex numbers? Real numbers? Integers? Commented Jan 5, 2020 at 23:30
• Complex matrices
– Dhdh
Commented Jan 5, 2020 at 23:32
• yes @dhdh, they are complex but if someone gave hints on real matrices, I can probably take it from there
– John
Commented Jan 5, 2020 at 23:36
• A conjugacy class is a set of matrices that are conjugate to a particular matrix. You wrote "a conjugacy class has to be equal to the identity". This doesn't make sense because a conjugacy class isn't a matrix; it is a set of matrices. There is one conjugacy class that contains the identity, namely $\{I\}$. The reason there are no other matrices in this set, is because the only matrix that is conjugate to the identity is the identity. Commented Jan 6, 2020 at 0:00

First type : $$\begin{equation*} \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & \frac{1}{ab} \end{pmatrix} \end{equation*}$$ for each nonzero $$(a,b)\in \mathbb{C}^2$$ and up to permutation of the diagonol terms. These are the diagonalisable cases.
Second type : $$\begin{equation*} \begin{pmatrix} a & 1 & 0 \\ 0 & a & 0 \\ 0 & 0 & \frac{1}{a} \end{pmatrix} \end{equation*}$$ for each nonzero $$a \in \mathbb{C}$$
Third type : $$\begin{equation*} \begin{pmatrix} a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & a \end{pmatrix} \end{equation*}$$ where $$a=1,j$$ or $$j^2 (j=e^\frac{2i\pi}{3})$$