Single $\text{GL}_n(\mathbb{C})$-conjugacy class, dimension as algebraic variety? I have two questions.


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*For each $c \in \mathbb{C}$, is the set$$\Sigma_c = \{A \in M_n(\mathbb{C}) : \text{Tr}(A) = c,\text{ }\text{Rank}(A) = 1\}$$a single $GL_n(\mathbb{C})$-conjugacy class?

*What is the dimension of $\Sigma_c$ as an algebraic variety over $\mathbb{C}$?


Many thanks in advance.
 A: (1) Hint The only Jordan block of rank $0$ is $\pmatrix{0}$, the only Jordan blocks of rank $1$ are $\pmatrix{c}$ and $$\pmatrix{0&1\\0&0},$$ and the rank of a direct sum of matrices is the sum of rank of the matrices, so the (positive) answer follows from the existence of the Jordan Canonical Form.
A: The dimension of $\Sigma_c$ is $2n-2$.
Proof. $\Sigma_c$ can be defined by $tr(A)=c$ and for every $2\times 2$ submatrix $B$ of$A$, $\det(B)=0$; then $\Sigma_c$ is an algebraic variety.
Case 1. $c\not= 0$. According to the Travis' post, $\Sigma_c$ is the conjugacy class of $Z=diag(c,0_{n-1})$. We consider the centralizer of $Z$ in $GL_n$: $C_Z=\{P\in GL_n;PZ=ZP\}$. The commutant of $Z$ is the vector space $diag(u,V_{n-1})$ -where $u,V$ are arbitrary- and has dimension $n^2-2n+2$. Since $C_Z$ is Zariski open dense in the commutant ($u\det(V)\not=0$), it has same dimension (as algebraic variety). Consequently $dim(\Sigma_c)=n^2-(n^2-2n+2)=2n-2$.
Case 2. $c=0$. Again according to Travis, $\Sigma_c$ is the conjugacy class of $Z=diag(U,0_{n-2})$ where $U=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. We consider the centralizer of $Z$ in $GL_n$: $C_Z=\{T\in GL_n;TZ=ZT\}$. We obtain $T=\begin{pmatrix}P&Q\\R&S\end{pmatrix}$ where $PU=UP,UQ=0,RU=0$ and $S$ is arbitrary. Then $P=aI_2+bU$, $Q$ has zero second line and $R$ has zero first column. The commutant of $Z$ has dimension $2+(n-2)+(n-2)+(n-2)^2=n^2-2n+2$ (as above). Since $\det(T)=a^2\det(S)$, $C_Z$ is a Zariski open dense in the commutant and has same dimension (as algebraic variety). We are done!
EDIT. A simpler proof for 2. Here $u,v$ are vectors.
$\Sigma_c=\cup_{i,j}C_{i,j}$, where $C_{i,j}=\{uv^T;u_i=1,v_j\not= 0,u^Tv=c\}$ (remark that, in $C_{i,j}$, $uv^T=u'v'^T$ implies $u=u',v=v'$). Note that $C_{i,j}$ is Zariski open dense in $D_{i}=\{uv^T;u_i=1,\sum_{p\not= i}u_pv_p+v_i=c\}$. Finally, the elements of $D_i$ depend on $2n$ parameters that are linked by $2$ algebraically independent relations; consequently $dim(D_i)=2n-2$ and we are done.
