Let $V$ be an inner product space over $\mathbb C$ and let $B = \{\vec u_1, \dots, \vec u_n\}$ be an orthonormal basis for $V$. Let $T \in \mathcal L(V, \mathbb F^n)$ and $S \in \mathcal L (\mathbb F^n, V)$ be operators given by $T(\vec x) = [\vec x]_B$ (coordinates of $\vec x$ in the basis $B$) and
$$S(\vec x) = \sum_{i=1}^n x_i \vec u_i$$
Show that $T^\ast = S$ i.e. $S$ is the adjoint of $T$.
My attempt: To show that $S = T^\ast$, we need to show that
$$ \langle T(\vec x), \vec y \rangle = \langle \vec x, S(\vec y) \rangle $$
For all $\vec x, \vec y \in \mathbb F^n$. Since $\beta$ is a basis, both $\vec x$ and $\vec y$ have a representation in it. Then
$$ \begin{align} \vec x &= a_1\vec u_1 + \dots + a_n\vec u_n \newline \vec y &= b_1\vec u_1 + \dots + b_n\vec u_n \end{align} $$
Taking $\langle T(\vec x), \vec y \rangle$, we have
$$ \begin{align}\langle T(\vec x), \vec y \rangle &= \langle [\vec x], \vec y \rangle \newline &= \langle [\vec x],b_1\vec u_1 + \dots + b_n\vec u_n \rangle \newline &= \overline{\langle b_1\vec u_1 + \dots + b_n\vec u_n,[\vec x] \rangle} \newline &= \sum_{i=1}^n \overline{b_i \langle \vec u_i, [\vec x] \rangle} \newline &= \sum_{i=1}^n \overline{b_i} \langle [\vec x], \vec u_i \rangle \end{align} $$
Where $[\vec x]$ is the coordinates of $\vec x$ in basis $B$ (For some reason typing [\vec x]_B doesn't work). Looking at the right-hand side,
\begin{align} \langle \vec x, S(\vec y) \rangle &= \langle \vec x, S(b_1\vec u_1 + \dots + b_n\vec u_n) \rangle \end{align}
This is where I am stuck. One idea I had is to use linearity
$$ \langle \vec x, S(b_1\vec u_1 + \dots + b_n\vec u_n) \rangle = \langle \vec x, S(b_1\vec u_1) + \dots + S(b_n\vec u_n) \rangle $$
But this didn't lead me anywhere, what would $b_iS(\vec u_i)$ be?
More generally, take two inner product spaces $V,W$ of dimension $n,$ an orthonormal basis $(u_1,\dots,u_n)$ for $V$ and an orthonormal basis $(w_1,\dots,w_n)$ for $W.$ This will apply to your $W=\Bbb F^n$ (where $\Bbb F=\Bbb R$ or $\Bbb C$), with its canonical inner product and orthonormal basis.
Let us check that the maps $T\in\mathcal L(V,W)$ defined by $$\forall i\in\{1,\dots,n\}\quad T(u_i)=w_i$$ and $S\in\mathcal L(W,V)$ defined by $$\forall i\in\{1,\dots,n\}\quad S(w_i)=u_i$$ are mutual adjoints, i.e. that $$S=T^*$$ otherly said $$T^{-1}=T^*,$$ i.e. let us check that $T$ is a unitary operator. But this is obvious, since it maps some orthonormal basis (of $V$) to an orthonormal basis (of $W$): for all $i,j\in\{1,\dots,n\},$ $$\langle T(u_i),w_j\rangle=\langle w_i,w_j\rangle=\langle u_i,u_j\rangle=\langle u_i,T^{-1}(w_j)\rangle$$ (where the two first terms are inner products in $W$ and the third and fourth terms are inner products in $V$).