Rank of limit points of a conjugacy class

Question

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

Answer 1

This has nothing to do with one-parameter subgroup. In general, a sequence of (possibly rectangular) constant-rank matrices may converge only to a matrix of equal or smaller rank.

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 $r<s$, 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 $\operatorname{rank}(Y)\le\liminf_{i\in\mathbb N}\operatorname{rank}(X_i)$.