Why is $L(\mathbb{R})$  just the set of multiplication maps? It makes sense that the only linear transformations on $\mathbb{R}$ would be maps of the form $x\mapsto ax$ for $a\in\mathbb{R}$. But how do we know these maps are all the possible linear transformations on the reals? Is there a way to prove that there isn't some weird function out there that just happens to be linear but isn't just a multiplication function?
 A: Suppose $T$ is a linear map from $\mathbb{R}$ to itself. Let $a = T(1)$. Then for any $x \in R$ we have $T(x) = T(x \cdot 1) = x \cdot T(1) = ax$ and so $T$ is just multiplication by $a$. So there is no weird function that is linear but isn't a multiplication function; any linear function from $\mathbb{R}$ to itself is a multiplication function.
A: Let $f: \mathbb R \to \mathbb R$ be a linear map. Note that $f$ is completely determined by the image of a basis, so in our particular case we have $f(x) = f(x\cdot 1) = x f(1)\quad$ for every $x \in \mathbb R$.
A: An $\mathbb{R}$-linear map $\mathbb{R}\to\mathbb{R}$ corresponds to a $1\times 1$-matrix. I challenge you to find $1\times 1$-matrices that don't have the form $(a_{11})$ for some scalar $a_{11}\in\mathbb{R}$ :) This is multiplication by $a_{11}$.
A: Note that $1\in\mathbb{R}$ is a "vector" and that $a\in\mathbb{R}$ is a "scalar" in the following computation: $f(a)=f(a\cdot 1)=af(1)$; the $\cdot$ denotes multiplication of a "scalar" by a "vector" (which, of course, is the same as multiplication of real numbers and hence the $\cdot$ is omitted in the entity after the final equality). If $c=f(1)\in\mathbb{R}$, then $f(a)=ca$ for all $a\in\mathbb{R}$ and thus $f$ equals multiplication by a fixed real number.
Exercise 1: Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$, then $f(z)=cz$ for all $z\in\mathbb{R}$ and some $c\in\mathbb{R}$. 
Exercise 2: If $f:\mathbb{R}\to\mathbb{R}$ is Lebesgue measurable and if $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$, then prove that $f$ is continuous. [Real and Complex Analysis (Second Edition) by Walter Rudin, Chapter 8, Exercise 24] 
Exercise 3: Prove (using the axiom of choice) that there exist real discontinuous functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. [Real and Complex Analysis (Second Edition) by Walter Rudin, Chapter 8, Exercise 24]
I hope this helps!
