Unraveling Theorem 4.2.5: Understanding Linear Maps and Scalar Products

[PcumP_Ravenclaw]

Theorem 4.2.5 The most general linear map $f : R^3 → R$ is of the form $x → x·a$, for some vector a in $R^3$.
Proof: Suppose that $f : R^3 → R$ is linear, and let $a = (a_1, a_2, a_3)$, where $a_1 = f (i), a_2 = f (j), a_3 = f (k).$
Then $f (x) = f (x_1i + x_2j + x_3k) = x_1 f (i) + x_2 f (j) + x_3 f (k) = x·a$.
My understanding
The function, f is scalar product and it takes two vectors x and a and changes them into a scalar x? why x again are they different? How did $f(x)$ become $f (x) = f (x_1i + x_2j + x_3k)$?

 

[Mark44]

Theorem 4.2.5 The most general linear map $f : R^3 → R$ is of the form $x → x·a$, for some vector a in $R^3$.
Proof: Suppose that $f : R^3 → R$ is linear, and let $a = (a_1, a_2, a_3)$, where $a_1 = f (i), a_2 = f (j), a_3 = f (k).$
Then $f (x) = f (x_1i + x_2j + x_3k) = x_1 f (i) + x_2 f (j) + x_3 f (k) = x·a$.
My understanding
The function, f is scalar product and it takes two vectors x and a and changes them into a scalar x?

No, f maps a vector $\vec{x}$ to the scalar $\vec{x} \cdot \vec{a}$. The dot (or scalar) product produces a scalar as its output.

why x again are they different? How did $f(x)$ become $f (x) = f (x_1i + x_2j + x_3k)$?

$f(\vec{i}) = \vec{i} \cdot \vec{a} = a_1$. Similar for $f(\vec{j})$ and $f(\vec{k})$.

 

[WWGD]

I think s/he is making use of some representation theorem for functionals in
$\mathbb R^n$ in which every linear functional can be described as the inner-product by a fixed element.