I'm working on the following problem in Royden:
Let $X$ be a finite dimensional normed linear space of dimension n. Let $(e_1, ..., e_n)$ be a basis for $X$. For $1 < i < n$, define $\phi_i(x)=x_i$ for: $$x = x_ie_i + +x_ne_n \in X (*)$$ Show that $(\phi_1, ... ,\phi_n)$ form a basis for for $X^*$. Thus dim $X^* = n$.
I'm pretty sure (*) should read $x = x_1e_1 + +x_ne_n \in X$ ($i \rightarrow 1$). I'm just making sure by asking the internet.
I'm able to argue spanning okay, but showing linear independence is not working for me. Suppose we can express any $\phi \in X^*$ as $\phi = \sum_{i=1}^n c_i \phi_i$. Then: $$0=0(x)=\sum_{i=1}^n c_i \phi_i(x)=\sum_{i=1}^n c_i x_i$$ I get suck here though because this doesn't imply all the $c_i$ are zero because the $x_i$ could also be zero. So how do I argue that all the $c_i$ should be zero? In another problem I saw that they evaluate $0$ on all the basis elements of $X$ to show that each coefficient had to be zero. I couldn't prove that was equivalent to what I was doing though.
Suppose you have a null linear combination of $\phi_i$ i.e $\sum_{i=1}^n \phi_i c_i=0$. Compute this in each basis element to get the result.