I'm going to give a partial answer here for two reasons: (1) I am lazy and (2) this is starting to feel a little homeworky to me. Obviously, no one would assign this material as homework, but part of reading a math paper is taking the time to work out lots of simple examples and see how the definitions work. I feel like you are pushing the boundaries of how much of this work it is reasonable to ask other people to do. Not a major criticism, certainly not a vote to close the question, but my input.

On to the math. I've scanned the first 3 pages of Hesselink's paper. He make the following definitions. G acts on V, v is a point of V and $\star$ a chosen base point of V fixed by G. In your setting, G is $GL_n$, V is the $n \times n$ matrices where G acts by conjugation, and $\star$ is zero. Hesselink writes Y(G) for what is essentially $\mathrm{Hom}(\mathbb{C}^*, G)$. More precisely, Hesselink tensors with $\mathbb{Q}$, so that he can talk about maps like $t \mapsto \left( \begin{smallmatrix} t^{1/3} & 0 \\\\ 0 & t^{-2/7} \end{smallmatrix} \right)$. I'll ignore this detail.

For $\lambda \in Y(G)$, Hesselink defines a rational number $m(\lambda)$. We talked about this in your previous question. In this setting, where V is an $N$-dimensional vector space, Hesselink gives an explicit formula for m on the bottom of page 142/top of page 143: Diagonalize the action of $\lambda$ as $t \mapsto \mathrm{diag}(t^{m_1}, \cdots, t^{m_N})$ and write $v = \sum v_i e_i$.. Then $m(\lambda) = \min(m_i : v_i \neq 0)$ if this number is nonnegative, and is $- \infty$ if this minimum is negative.

Let's see what this definition means in your setting. We can conjugate any $\lambda$ into diagonal form as $t \mapsto \mathrm{diag}(t^{c_1}, \cdots, t^{c_n})$. I've replaced $m_i$ by $c_i$ to point out that these $c$'s are not the $m$'s of the previous paragraph. In our notation, the $N$ of the previous paragraph is $n^2$. The vector space $V$ has dimension $n^2$ with basis $e_{ij}$. The action of $\lambda(t)$ on $e_{ij}$ is by $t^{c_i - c_j}$. (**Exercise!**).

So $m(\lambda) > 0$ if and only if $c_i \leq c_j$ implies $v_{ij} =0$.

We may as well order our basis such that $c_1 \geq c_2 \geq \cdots \geq c_n$.
If $c_1 > c_2 > \cdots >c_n$ then we see that $m(\lambda) > 0$ if and only if $v$ is a strictly upper triangular matrix. When there are some equalities among the $c$'s, you want $v$ to be strictly block upper triangular. For such a $v$, $m(\lambda) = \min(c_i - c_j : v_{ij} \neq 0)$. In particular, notice that there exists a $\lambda$ such that $m(\lambda) > 0$ if and only if $v$ is nilpotent.

Hesselink defines $\Lambda(v)$ to be the locus in $\{ \lambda : m(\lambda) = 1 \}$ where $q(\lambda)$ is minimized, where $q$ is the inner product from your previous question. What you want to show is that $\Lambda(v)$ determines the Jordan normal form of $v$.

I must admit that I haven't thought out how to prove this. But I hope this makes things explicit enough that you can attack it.