If $u=\sum_{i=1}^kL_i\otimes x_i\in\mathcal L(X,Y)\otimes X$, is $T(u):=\sum_{i=1}^kL_i(x_i)$ independent of the representation of $u$?

Question

Let $X,Y$ be $\mathbb R$-vector spaces and $u\in\mathcal L(X,Y)\otimes X$ with $$u=\sum_{i=1}^kL_i\otimes x_i\tag1$$ for some $k\in\mathbb N$, $L_1,\ldots,L_k\in\mathcal L(X,Y)$ and $x_1,\ldots,x_k\in X$. Are we able to show that the value $$T(u):=\sum_{i=1}^kL_i(x_i)$$ is independt of the representation $(1)$ of $u$? I know how this can be proved for $Y=\mathbb R$, but don't know how we can argue in general.

Answer 1

Hint : define $T: \mathcal{L}(X,Y)\times X \to Y$ by $T(L,x) = L(x)$. Prove that this is bilinear, then use the universal property of the tensor product.