In Linear Algebra Done Right we have a theorem that states
Riesz Representation Theorem : Suppose $V$ is finite-dimensional and $A$ is a linear functional on $V$. Then there is a unique vector $u$ such that for every $v$ : $ A(v)=\langle u , v\rangle$
However then the existence of adjoint transformation is cited using this theorem
$\langle T v, w\rangle=\left\langle v, T^{*} w\right\rangle$
To see why the definition above makes sense, suppose $T \in \mathcal{L}(V, W)$. Fix $w \in W$. Consider the linear functional on $V$ that maps $v \in V \text { to }\langle T v, w\rangle$; this linear functional depends on $T$ and $w$. By the Riesz Representation Theorem , there exists a unique vector in V such that this linear functional is given by taking the inner product with it. We call this unique vector $T^{*} w$
I don't understand how the two situations are alike. First we didn't have a transformation prior to inner product and now we do. How does the guarantee still exist? Not only that but $v$ and $w$ could belong to different dimension spaces and $T$ transforms $V$ to $W$. How does Riesz Representation Theorem hold for this case at well? The two seem a bit disconnected to me.
Riesz representation theorem does not mention transforming input by T. By applying this theorem exactly as it is stated, there exists a $w$ for the functional acting on $Tv$, so now we have $⟨,⟩$. I don't see ahead of that.
Suppose $$T_v(x)=\langle x,v\rangle$$ Given $y \in V$. Consider the linear functional $$g:V\to\mathbb{F}$$ by $$g(x)=\langle T(x),y\rangle$$ for any $x\in V$
Note that $g$ depends on $y$.
Since $g \in \mathcal{L}(V,\mathbb{F})$, by Riesz Representation Theorem $$g(x)=\langle x,y'\rangle=T_{y'}(x)$$ where $y' \in V$ is unique.
i.e. for any $x \in V$ $$\langle T(x),y\rangle = g(x)=T_{y'}(x)=\langle x,y'\rangle$$ We want it to be equal to $\langle x,T^*(y)\rangle$
Therefore, we define $$T^*:V\to V$$ by$$T^*(y)=y'$$
After checking $T^*$ is linear, the existence of $T^*$ has been shown.