Why is $\mathbb{C}$ a vector space over $\mathbb{R}$ but $\mathbb{R}$ not a vector space over $\mathbb{C}$?

Question

My book (Álgebra Linear by Isabel Cabral, Cecilia Perdigao and Carlos Saiago) asks the question in the title but doesn't offer solutions. My guess is that the elements of $\mathbb{R}$ are in $\mathbb{C}$ but not all the elements of $\mathbb{C}$ are in $\mathbb{R}$, so you can't do any operations on it, and a vector space is defined by a set and operations on that set? Would this set have "gaps" or would it be empty, because every element of $\mathbb{C}$ can be represented as $a+bi$ and $\mathbb{R}$ doesn't have complex numbers?

Answer 1

In order to define in $\mathbb{R}$ a structure of $\mathbb{C}$-vector space, you must define an operation (called scalar multiplication) $$(-*-)\colon\mathbb{C}\times \mathbb{R}\rightarrow\mathbb{R}$$ $$(z,v)\mapsto z*v$$ But the obvious operation (i.e, the product in $\mathbb{C}$) doesn't work, as you said, because $zv$ does not have to be in $\mathbb{R}.$

You can prove that in fact you can't define any operation which respects the conditions of vector space that induces the usual multiplication when restricted to $\mathbb{R}\times\mathbb{R}:$

Let's say that if $z\in\mathbb{R}$ then $z*v=zv$, the usual multiplication in $\mathbb{R}.$ Then $(-*-)$ is determined by $(\textbf{i}*1)$, because every $z\in\mathbb{C}$ can be written as $a+b\textbf{i},\ a,b\in\mathbb{R},$ so for every $v\in\mathbb{R}$ $$(a+b\textbf{i})*v=a*v+b\textbf{i}*v=av+bv(\textbf{i}*1)$$ (I'm using here the distributive property of a vector space). But, if $(\textbf{i}*1)=a,$ then $(a-\textbf{i})*1=a*1-\textbf{i}*1=a-a=0$, which is a contradiction since $a\neq\textbf{i}$ and $ 1\neq 0$.