Consider a linear transformation $\mathsf T$ from $\Bbb R^3$ to $\Bbb R$. Show that there exists a vector $\vec v$ in $\Bbb R^3$ such that $\mathsf T(\vec x)$ = $\vec v \cdot \vec x$, for all $\vec x$ in $\Bbb R^3$.
I don't understand how the dot product is a linear transformation considering that the output is a scalar. And I have no idea how to prove this.Can someone shed some light for me.
Let ${\mathsf T}(\vec e_1)=a$, ${\mathsf T}(\vec e_2)=b$, and ${\mathsf T}(\vec e_3)=c$, then by setting $\vec v=(a,b,c)$, we see that for any $\vec x=(x_1,x_2,x_3)\in\mathbb{R}^3$, \begin{align} {\mathsf T}(\vec x) &={\mathsf T}(x_1\vec e_1+x_2\vec e_2+x_3\vec e_3)\\ &=x_1{\mathsf T}(\vec e_1)+x_2{\mathsf T}(\vec e_2)+x_3{\mathsf T}(\vec e_3)\\ &=a\,x_1+b\,x_2+c\,x_3\\ &=\vec v\cdot\vec x. \end{align} This completes the proof.
Consider the zero transformation function as $T: \mathbb{R^3} \rightarrow{} \mathbb{R}$. Then it is trivial that $$T(\vec{x}) = \vec{0} \cdot \vec{x}$$
