Let $H$ be a hypergraph on a finite vertex set. Prove that $\nu^*(H)=\tau^*(H)$ where $\nu^*(H)$ is the fractional matching number and $\tau^*(H)$ is the fractional transversal number of $H$.

Let $H$ be a hypergraph on a finite vertex set. Prove that $\nu^*(H)=\tau^*(H)$ where $\nu^*(H)$ is the fractional matching number and $\tau^*(H)$ is the fractional transversal number of $H$.

Let $H=(V,E)$ be a hypergraph where $V$ is a set of vertices and $E$ is a set of hyperedges.
Let $A$ be an incidence matrix of $H$ with its size $m\times n$ with $m := |E|, n:=|V|$ such that $A_{ij} \in \{ 0, 1\}$, $A_{ij}=1$ if i-th edge contains j-th vertex, $A_{ij}=0$ else.
By definition of $\tau^*$ and $\nu^*$, $\tau^*(H)=min\{ \mathbf{1}_n^T x : x \in P\}$ and $\nu^*(H)=max \{ y^T \mathbf{1}_m : y \in D\}$, where $P=\{ x \in \mathbb{R}^n : x\geq 0 , Ax \geq \mathbf{1}_m \}$ and $D=\{ y\in \mathbb{R}^m : y\geq 0, y^TA \leq \mathbf{1}_n^T\}$.
Trivially, $0 \in D$.
By the definition of hyperedge, $A \mathbf{1}_n \geq \mathbf{1}_m$, so $\mathbf{1}_n \in P$.
Therefore, $P\neq \emptyset$ and $Q \neq \emptyset$.
By Duality of linear programming, $\tau^*(H) = \nu^*(H)$.