If W is a subspace of V and $x \notin W$, prove that there exists $f \in W^0$ such that $f(x) \neq 0$.

Question

Let V be any finite dimensional vector space over F. For any subset S of V, the annihilator$ S^0$ of S is the set $S^0=\{f \in V^* | f(x)=0 ,\forall x \in S\}$. If W is a subspace of V and $x \notin W$, prove that $ \exists f \in W^0$ such that $f(x) \neq 0$. (here,$ V^* $means V dual space)

My question is does this violate the definition of annihilator, which says that "annihilator$ S^0$ of S is the set $S^0=\{f \in V^* | f(x)=0 ,\forall x \in S\}$? Also, how am I supposed to prove the existence? Am I supposed to use replacement theorem here to extend the basis of $W?$

Answer 1

There's only a violation if $x \subsetneq W$ but since the question actually says $ x \notin W$, there's no violation. From there, the existence of $x$ is almost consequential (if there didn't exist such an $x$ then $W$ would have to be the whole space $V$ and not just a subspace i.e. $W^0$ would be the zero map which is the dual space of $V$)

Answer 2

In this problem, $x \notin W$ is a nonzero vector in $V,$ since $0 \in W.$

Let $\{ v_{1},v_{2},\dots,v_{k}\}$ be a basis for $W.$ Then, $\{ v_{1},v_{2},\dots,v_{k}\} \cup \{x\} = \{ v_{1},v_{2},\dots,v_{k}, x\}$ is a linearly independent subset of $V,$ since $x \notin span(\{ v_{1},v_{2},\dots,v_{k}\}).$

Now, extend this linearly independent subset of $V$ to a basis $\{ v_{1},v_{2},\dots,v_{k},v_{k+1},\dots,v_{n}\}$ for $V,$ where $v_{k+1}=x$ for notational convenience. Then, there is an ordered basis $\{ f_{1},f_{2},\dots,f_{k},f_{k+1},\dots,f_{n}\}$ for the dual space such that it is the dual basis for $\{ v_{1},v_{2},\dots,v_{k},v_{k+1},\dots,v_{n}\}.$

In turn, there is a linear functional $f_{k+1}$ in the dual basis such that $f_{k+1}(x) = f_{k+1}(v_{k+1}) = 1 \neq 0$ and $f_{k+1}(v_{i})=0,$ where $i=1,2,\dots,k.$ Thus, there is a linear functional $f=f_{k+1} \in W^{0}$ with the desired property.