Suppose $V $ is a finite dimensional vector space and $U $ is a subspace of $V$. Show that if $x\notin U$ then there exists an $a \in V^*$ such that $a(x) = 1$ and $a(u) = 0 $ for all $u \in U$. Note that $V^*$ is the dual space of V.
I think that taking the basis of $U$ will allow me to construct a case for all $a(u) = 0 \in U$ as the basis of $U$ is linearly independent. But how do I go about showing an $x \notin U$ such that $a(x) = 1$ ?
I am fairly new to proofs and dual spaces so I might have my logic completely muddled up.
I would be grateful for any help with the proof. Thank you in advance.
A linear functional is completely defined by its behavior on a basis, so just extend your basis for $U$ to one for $U+span(x)$ and define $a$ to be $0$ on all of $U$ and $1$ on $x$.