# Rank of limit points of a conjugacy class

Problem: Let $$t \to P_t$$ be a one parameter subgroup $$\mathbb{C}^* \to \text{Gl}_{n}(\mathbb{C})$$.

Let $$X$$ be a $$n \times n$$ nilpotent matrix.

I want to show that if $$lim_{t \to 0} P_t^{-1}XP_t = Y$$ then $$\text{rank}(X) \geq \text{rank}(Y)$$.

I have checked that this is true if $$X$$ is in the Jordan canonical form and the one parameter subgroups are in diagonal form. Firstly in this case we can assume that $$X$$ is indecomposable(hence rank is n-1) and so only one jordan block(since the actions are blockwise due to the nature of $$P_t$$), then if $$P_t(i, i) = t^{i}$$, then the $$Y$$ as defined above has $$t^{i_k +i_{k+1}}$$ on the super diagonal entries and $$0$$ elsewhere.

So, the rank is at most $$n-1$$.

But for an arbitrary nilpotent matrix I don't know how to show.

We can make one simplification: if $$Z$$ is a nilpotent matrix then $$Z = A^{-1}XA$$ where $$X$$ is in Jordan canonical form, but then we cannot assume that $$P_t$$ are diagonal and so on...

So, either we prove that given $$X$$ in Jordan Canonical form and $$P_t$$ in arbitrary form..

or $$X$$ in arbitrary form and $$P_t$$ in special form that the rank cannot increase on taking limits.

I am interested in this question since combined with

1. fact that any point in the Zariski closure of a conjugacy class in the affine variety of these nilpotent matrices is captured by a one parameter subgroup

2. If $$Y$$ is in closure of $$X$$ then, $$Y^2$$ is in closure of $$X^2$$ and so on....

3. Partition(there is one partition corresponding ot every class) of $$X$$ majorizes partition of $$Y$$ iff $$rank^{i}(X) \geq rank^{i}(Y)$$

we would have shown that if $$Y$$ is in the closure of the conjugacy class of $$X$$ then the partition corresponding to $$X$$ majorizes that of $$Y$$.

I have shown the converse using one parameter subgroups and block matrix identitities here

Let $$Y=\lim_{i\to\infty}X_i$$, where each $$\operatorname{rank}(X_i)=r$$ for each $$i$$ and $$\operatorname{rank}(Y)=s$$. If $$Y=0$$, there is nothing to prove. Suppose $$Y\ne0$$. Perform an economic singular value decomposition $$Y=U\Sigma V^\ast$$ where $$V$$ has $$s$$ orthonormal columns. Let $$\mathcal V$$ be the column space of $$V$$. Then $$\|Yv\|_2\ge\sigma_s(Y)>0$$ for all unit vectors $$v\in\mathcal V$$.
If $$r, then $$n<(n-r)+s$$. Hence $$\ker(X_i)\cap\mathcal V\ne0$$ and there exists some unit vector $$u_i\in \ker(X_i)\cap\mathcal V$$. Since the intersection of the unit sphere and $$\mathcal V$$ is compact, $$\{u_i\}$$ has a subsequence $$\{u_{i_k}\}_{k\in\mathbb N}$$ that converges to some unit vector $$v\in\mathcal V$$. But then $$Yv=\lim_{k\to\infty}X_{i_k}u_{i_k}=0$$, which contradicts our previous observation that $$\|Yv\|_2>0$$. Hence we must have $$r\ge s$$.
Remark. By considering an appropriate subsequence, the above result is equivalent to the following: if $$X_1,X_2,\ldots$$ have possibly different ranks but they converge to $$Y$$, then $$\operatorname{rank}(Y)\le\liminf_{i\in\mathbb N}\operatorname{rank}(X_i)$$.