Let $V$ be a finite dimensional inner product space with inner product $\langle , \rangle$. Show that for any $f \in V^*$, there exists a unique vector $w \in V$ so that $f(v)=\langle w,v \rangle$ for all $v \in V$.
Uniqueness.
Suppose there exists $w_1, w_2 \in V$ such that $f(v) = \langle w_1, v \rangle = \langle w_2, v \rangle$. Therefore, $\langle w_1 - w_2, v \rangle = 0$ for all $v \in V$. Taking $v = w_1 - w_2$ gives $\langle w_1 - w_2, w_1 - w_2 \rangle = 0$. Since $\langle, \rangle$ is an inner product this means that $w_1 = w_2$.
I'm not sure how to prove existence, though. How do I do this?
This is a very standard theorem in linear algebra, which is also true for more general Hilbert spaces. See for instance Axler's Linear Algebra Done Right.
In the proof, the following is used: