Let $V$ be an $\mathbb R$ - vector space and let $V^* = Hom _\mathbb R (V, \mathbb R)$ be the dual space of $V$.
Define $f: V^* \times V \to \mathbb R$ by $$f(l^*, \vec v)=l^*(\vec v) \text{ for all } (l^*, \vec v) \in V^* \times V$$ is a bilinear map. Use the universal mapping property for tensor product to prove that there exists a well-defined linear map $T: V^* \otimes_\mathbb R V \to \mathbb R$ defined by $$ T (c_1l_1^* \otimes \vec v_1+ \dots+c_kl_k^* \otimes \vec v_k)=c_1l_1^*(\vec v_1)+ \dots + c_kl_k^*(\vec v_k)$$ for $c_1,\dots,c_k \in \mathbb R$, $l_1^*, \dots, l_k^* \in V^*$ and $\vec v_1, \dots, \vec v_k \in V$.
Solution: Given $f: V^* \times V \to \mathbb R$ is a bilinear map. Then by definition of a tensor product, there exists a bilinear map $\pi: V^* \times V \to V^* \otimes_\mathbb R V$ satisfying the following universal property: Given any bilinear map $f$, there exists a unique linear map $T$ such that $T \circ \pi=f$. Therefore, $T: V^* \otimes_\mathbb R V \to \mathbb R $ defined by $ T (c_1l_1^* \otimes \vec v_1+ \dots+c_kl_k^* \otimes \vec v_k)=c_1l_1^*(\vec v_1)+ \dots + c_kl_k^*(\vec v_k)$ for all $c_1,\dots,c_k \in \mathbb R$, $l_1^*, \dots, l_k^* \in V^*$ and $\vec v_1, \dots, \vec v_k \in V$ exits and is well-defined.