Linear Functional on Dual Space

Question

Before I state the problem, first recall that for $x \in V$, $\hat{x} : V^* \to F$ is defined by $\hat{x}(f) = f(x)$ for every $f \in V^*$, where $V^*$ is the dual space of $V$.

Let $V$ be a finite dimensional vector space, and let $x \in V$. If $\hat{x}(f)=0$ for all $f \in V^*$, then $x=0$.

Let $\beta$ be a basis for $V$, and let $\beta^* = \{f_1,..,f_n\}$ be its dual basis. Then $\hat{x}(f_i) = 0$ or $f_i(x)=0$ for every $i$. This means that $[x]_{\beta} = (0,0,...,0)^T$, and since $x \mapsto [x]_{\beta}$ is an isomorphism, $x$ must be $0$ since the kernel is trivial.

How does this sound?

Answer 1

Sounds good. But we can avoid the isomorphism and prove the result in infinite dimension too, as follows:

Let $V$ be a vector space and $x \in V$ such that $f(x) = 0$ for all linear maps $f\colon V \to K$. Then $x=0$.

Assume not. Then $\{x\}$ is linearly independent and so we can take a basis $B = \{ x_i \}_{i \in I}$ with $x \in B$. Define $f\colon B \to K$ putting $f(x_i) = 1$ for all $i \in I$. We have an unique linear extension (also denoted by $f$), $f\colon V \to K$. Then $f(x) = 1 \neq 0$. Done.