Existence of adjoint via Riesz Representation Theorem

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 $$⟨𝑇𝑣,𝑤⟩$$. I don't see ahead of that.

• I don't see what the issue is. For the Riesz representation theorem, you have a linear functional $A : V \to \mathbb{R}$ and deduce $A$ takes the form $Av = \langle u,v \rangle$. In the other setting, you have $A: V \to \mathbb{R}$ (given by $Av = \langle Tv,w \rangle$). So, applying, the Riesz representation theorem, there is some $u$ for which $Av = \langle u,v \rangle$. We denote $u$ by $T^*w$. – mathworker21 Jul 20 at 4:34
• Riesz representation theorem does not mention transforming input by T. By applying this thorem exactly as it is stated, there exists a $w$ for the functional acting on $Tv$, so now we have ⟨𝑇𝑣,𝑤⟩. I dont see ahead of that – Rahul Deora Jul 20 at 4:46
• Dude, read my previous comment. Define the linear functional $A : V \to \mathbb{R}$ as follows. For a given vector $v \in V$, let $Av = \langle Tv, w\rangle$. This is a valid linear functional, so Riesz representation theorem implies there is some $u$ for which $Av = \langle u,v\rangle$ for each $v \in V$. By the definition of $A$ this means $\langle Tv,w\rangle = \langle u,v \rangle$ for each $v \in V$. What exactly do you have a problem with? Be specific. – mathworker21 Jul 20 at 4:48
• I'm confused about how the T got stuck in there without the theorem changing. My defination of linear functional only extends to a dot product of two vectors as 𝐴𝑣=⟨𝑢,𝑣⟩ – Rahul Deora Jul 20 at 5:15
• With that definition of linear functional, the riesz representation theorem you cited is a tautology. Your definition of linear functional is obviously the wrong one. – mathworker21 Jul 20 at 5:31

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.