I am currently stuck when trying to show a seemingly trivial equation.
Let $X$ be a vector space over $\mathbb{R}$, $n\in\mathbb{N}$ and $m:\left\{1,2,\ldots n\right\}\rightarrow\mathbb{N}.$
For $i = 1,2,\ldots n$ and $j\leq m(i)$ we have scalars $\lambda_{ij} \in \mathbb{R}^{+}$ and vectors $x_{ij} \in X$.
I want to show that every $x:=\sum_{i=1}^{n} \sum_{j=1}^{m(i)}\lambda_{ij}x_{ij}$ can be written as
$$x = \sum_{j_1= 1}^{m(1)} \cdots \sum_{j_n= 1}^{m(n)} \mu_{j_{1}\cdots j_{n}} \sum_{i=1}^{n} (\sum_{j=1}^{m(i)} \lambda_{ij}) x_{ij_{i}}$$
with $\mu_{j_{1}\cdots j_{n}} \in \mathbb{R}^+$ and $\sum_{}^{}\mu_{j_{1}\cdots j_{n}} = 1$.
I read this once and wondered how it is possible to chose such a convex combination. I tried everything but couldn't come up with a way to find those $\mu_{j_{1}\cdots j_{n}}$.
It works with $$ \mu_{j_{1} \cdots j_{n}} := \prod_{i=1}^{n} \frac{\lambda_{ij_{i}}}{\sum_{j=1}^{m(i)} \lambda_{ij}}. $$