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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 $βŸ¨π‘‡π‘£,π‘€βŸ©$. I don't see ahead of that.

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    $\begingroup$ 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 $T^*w$. $\endgroup$ – mathworker21 Jul 20 '19 at 4:34
  • $\begingroup$ 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 βŸ¨π‘‡π‘£,π‘€βŸ©. I dont see ahead of that $\endgroup$ – Rahul Deora Jul 20 '19 at 4:46
  • $\begingroup$ 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. $\endgroup$ – mathworker21 Jul 20 '19 at 4:48
  • $\begingroup$ 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=βŸ¨π‘’,π‘£βŸ© $\endgroup$ – Rahul Deora Jul 20 '19 at 5:15
  • $\begingroup$ 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. $\endgroup$ – mathworker21 Jul 20 '19 at 5:31
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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.

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