I've recently got some homework to do, such as this exercise:
Let $X$ be a $\mathbb{K}$-finite vector space. Show that $X^*=\{\phi :X\longrightarrow \mathbb{K}:\phi\text{ is linear}\}$ is a vector space. Let $\{b_1,b_2,\ldots,b_n\}$ a basis of $X$. Show that $\phi_j(b_k):=\delta_{j,k}$ for all $j,k$ is a well-defined linear map $\phi_j:X\longrightarrow\mathbb{K}$. Show that $\{\phi_1,\ldots,\phi_n\}$ is a basis of $X^*$.
So I've done the first part, but I've tried everything to prove that the $\phi_j$'s are linear without success ! I really don't have a clue anymore (I tried the basic linear properties, tried to make the map as a composition of linear ones etc...)
Help please, and thank you
Suppose that $x_1\phi_1+x_2\phi_2+\cdots+x_n\phi_n=0$. Then$$(x_1\phi_1+x_2\phi_2+\cdots+x_n\phi_n)(b_1)=0,$$but this means that $x_1=0$, since$$(x_1\phi_1+x_2\phi_2+\cdots+x_n\phi_n)(b_1)=x_1\phi_1(b_1)+x_2\phi_2(b_1)+\cdots+x_n\phi_n(b_1)=x_1.$$By the same argument, each $x_k$ is $0$. Therefore, $\{\phi_1,\phi_2,\ldots,\phi_n\}$ is linearly independent.
The $\phi_j$'s are linear because $\{b_1,b_2,\ldots,b_n\}$ is a basis, $\mathbb K$ is a vector space and therefore there is one and only one linear map from $X$ into $\mathbb K$ such that, for each $k$ and each $j$, $\phi_j(b_k)=\delta_{j,k}$.