This question comes from Exercise 4.1 of Lectures of geometric constructions, Kamnitzer - arXiv link, and is a follow on of a question I posted here which deals with part 1 of the exercise. This question is on part 2.
In this exercise, we are given that $X: \mathbb{C}^N \to \mathbb{C}^N$ is a nilpotent matrix with $X^n = 0$. Associated to it is the partition $ \mu = (\mu_1, \dots, \mu_n) $ with $$\mu_i = \dim \ker (X^i) - \dim\ker(X^{i-1}).$$
To $X$ we can associate also associate the partition $\nu = (\nu_1, \dots, \nu_m)$ where each $\nu_i$ is the size of the $i$-th Jordan block of $X$, imposing an ordering so the 1st Jordan block is of largest size, and so on. The Young diagram of $\nu$ has a conjugate partition $\lambda$, where each $\lambda_i$ is the number of $j$ such that $\nu_j \geq i$ (i.e. it is the number of Jordan blocks of size greater than or equal to $i$).
The first part shows that for all $k$, $$\mu_1 + \dots + \mu_k \leq \lambda_1 + \dots + \lambda_k.$$ Now I need to show that as $GL_n$ weights, $\lambda \geq \mu$.
We have $$ \lambda - \mu = ((\lambda_1-\mu_1), \dots, (\lambda_n-\mu_n)),$$ which I want to express as a sum of $$k_1 a_1 + \dots + k_{n-1}a_{n-1},$$ where $k_i$ are non-negative integers and $a_1 = (1,-1,0,\dots,0), \dots, a_{n-1} = (0, \dots, 0, 1, -1)$. In other words, the setup becomes $$((\lambda_1-\mu_1), \dots, (\lambda_n-\mu_n)) = (k_1, k_2-k_1, \dots, k_{n-1}-k_{n-2}, -k_{n-1}).$$
Starting from the left, we get an inductive equation $$k_i = (\lambda_1 + \dots + \lambda_i) - (\mu_1 + \dots + \mu_i),$$ but I am unable to show that $$-k_{n-1} = \lambda_n-\mu_n \tag{*}.$$ Actually, thanks to Joppy's answer, I understand now, just rearrange the terms by using the fact that both $\mu$ and $\lambda$ are partitions of $N$. I'm pretty sure my calculations are correct, and the only place where we use the part 1 inequality is to show that the constants $k_i$ are non-negative, but that's all fine - the only part that I am stuck on is showing (*).
The weight space of the algebraic group $\operatorname{GL}_n$ is $X = \mathbb{Z}\{\epsilon_1, \ldots, \epsilon_n\}$. The simple roots are $\alpha_1, \ldots, \alpha_{n - 1}$ where $\alpha_i = \epsilon_i - \epsilon_{i + 1}$. We say that for two weights $\mu, \lambda \in X$ that $\mu \leq \lambda$ if and only if $\lambda - \mu = \sum_{i = 1}^{n-1} m_i \alpha_i$ for $m_i \geq 0$.
Claim: Writing $\mu = \sum_{i = 1}^n \mu_i \epsilon_i$ and $\lambda = \sum_{i = 1}^n \lambda_i \epsilon_i$, we have that $\mu \leq \lambda$ if and only if $\mu_1 + \cdots + \mu_i \leq \lambda_1 + \cdots + \lambda_i$ for all $i \in \{1, \ldots, n - 1\}$ and $\mu_1 + \cdots + \mu_n = \lambda_1 + \cdots + \lambda_n$.
Proof: Suppose $\mu \leq \lambda$ with $\lambda = \mu + \sum_{i = 1}^{n - 1} m_i \alpha_i$. For any $1 \leq i \leq n - 1$, we can take the sum of the $\epsilon_1, \ldots, \epsilon_i$ coefficients on both sides to get $\lambda_1 + \cdots + \lambda_i = \mu_1 + \cdots + \mu_i + m_i$, so the inequality follows because of the weak positivity of the $m_i$. Summing all coefficients gives $\mu_1 + \cdots + \mu_n = \lambda_1 + \cdots + \lambda_n$. It's clear that this is both necessary and sufficient, since we can reconstruct the $m_i$ from the differences $m_i = (\lambda_1 + \cdots + \lambda_i) - (\mu_1 + \cdots + \mu_i)$.
So, somewhere you really do need to use the fact that the partitions are the same size, since partitions of different sizes are never related under $\leq$.