conjugacy classes of a matrix satisfies an equation I met with a linear algebra problem which ask me to describe all conjugacy classes of $6\times6$ matrices $A$ which satisfy the equation $(A+1)^2(A-1)^4=0$.
Since I am not very familiar with abstract algebra, I think I do not quite understand the conjugacy classes(for a matrix).
I searched the definition, and in abstract algebra, conjugate classes of $a$ in a group $G$ is $\{xax^{-1}|x \text{ in } G\}$. So how does this related to a matrix? All the matrices similar to $A$(as the conjugate class of A)?
How could I use the condition $(A+1)^2(A-1)^4=0$? Since the dimension of A is 6, does it mean it should have two root $-1$, and four root $1$? And thus just consider all possible ways of Jordan blocks is OK?
Thank you very much!  
 A: The question is to classify up to similarity (base change) all linear operators on a $6$-dimensional space whose minimal polynomial divides $(X+1)^2(X-1)^4$. Nothing is specified about the field, but I will suppose it is not of characteristic$~2$ (in which case that polynomial would become $(X+1)^6$) which is the only relevant point here.
Since the minimal polynomial is split with roots $-1,1$ only, the operator admits a matrix in Jordan normal form, with Jordan blocs for $\lambda\in\{-1,1\}$ only. Moreover the divisibility condition says that Jordan blocs for $\lambda=-1$ cannot have size exceeding$~2$, and those for$\lambda=1$ cannot have size exceeding$~4$. You therefore get a fairly long list of possibilities, obtained as follows: write $6=k+l$ in all $7$ possible ways, and then for each combine all partitions of $k$ into parts of size at most$~2$ with all partition of$~l$ into parts of size at most$~4$.
The length of the list is$~44$: this is the coefficient of $X^6$ in $\frac1{(1-X)(1-X^2)}\times\frac1{(1-X)(1-X^2)(1-X^3)(1-X^4)}$ evaluated as power series.
