I'm not sure if my solution for the following problem is valid.
Let $(V, ( , ))$ be a finite-dimensional inner product space over $F$ and let $V^*$ denote its dual space. Prove that for all $l \in V^*$, there exists a unique $w \in V$ such that: $$l(\vec{v})=(\vec{v},\vec{w}), \forall \vec{v}\in V$$
My solution versus one by the prof. :
The solution given by the instructor in the course I'm taking basicly consists of constructing an orthonormal basis $(u_j)$ and then writing $l(\vec{v})$ as a sum of $(v,u_j)l(u_j)$, which can be rewritten as a unique inner product proving existence of the desired $w$. Uniqueness follows immediately.
My attempt to solve the problem was the following, first I let $\alpha=\{\vec{u_1},\cdots,\vec{u_n}\}$ be a basis for $V$. Then, the matrix representation for $l\in V^*$ w.r.t $\alpha$ and the basis (1) for $F$ is given by: $$\bigg[l(\vec{u_n})\cdots l(\vec{u_n})\bigg]$$ But then we have that $l(\vec{v})=\bigg[l(\vec{u_1})\cdots l(\vec{u_n})\bigg]\vec{v}$ So, taking $\vec{w}=\bigg[l(\vec{u_1})\cdots l(\vec{u_n})\bigg]^*$ we would have the desired property that $l(\vec{v})=(\vec{v},\vec{w})$. Again, uniqueness would follow immediately. Obs: $^*$ denotes transpose and conjugate, i.e adjoint.
What is wrong with my proof?