On showing that all linear maps $T: \mathrm{F}^n \to \mathrm{F}$ are inner products, $\mathrm{F} \in \{\mathrm{R}, \mathrm{C}\}$ My reference book claims that all linear maps $T: \mathrm{F}^n \to \mathrm{F}^m$ are given by matrices, and thus by inner products. While it is quite trivial to show that any inner product is in fact a linear map $T: \mathrm{F}^n \to \mathrm{F}$, for $\mathrm{F} \in \{\mathrm{R}, \mathrm{C}\}$, I'm not sure how to proceed to other way around, namely how do you argue that there cannot be any other linear mapping $T': \mathrm{F}^n \to \mathrm{F}$, which does not have the form $T(x) = \left<x, a\right>, a \in \mathrm{F}^n$, when the given properties of linear mappings are that:

Let $V$ and $W$ be vector spaces over the field $F$. Then a function $T: V \to W$ is said to be a linear mapping if i.) $\forall x, y \in V: T(x + y) = T(x) + T(y)$, ii.) $\forall \lambda \in \mathrm{F}: \forall x \in V: T(\lambda x) = \lambda T(x)$


Let $e_1,\dots,e_n$ denote the standard basis of $F^n$ (so for instance, $(0,1,0,\dots,0)$). For any linear map $T:F^n \to F$, we find that for any $x = (x_1,\dots,x_n) \in F^n$, we have
$$
T(x) = T(x_1 e_1 + \cdots + x_n e_n) = 
x_1T(e_1) + \cdots + x_n T(e_n).
$$
In other words, $T$ is the dot-product of $x$ with the vector $(T(e_1),\dots,T(e_n))$. So, every linear map $T$ from $F^n$ to $F$ can be expressed as a dot-product.
