Intuition behind why $\mathbb{C} \cong \mathbb{R}^2$

Question

We learned in my linear algebra course last semester that two vectors spaces $U$ and $V$ over some field $F$ are said to be ${\it isomorphic}$ if there exists a bijection (invertible linear map) between them.

Consider the vector spaces $\mathbb{C}$ and $\mathbb{R}^2$. If we consider $\mathbb{C}$ as

$$ \mathbb{C} = \left\{ a + bi \ | \ a, b \in \mathbb{R} \right\}, $$

then it seems that $\mathbb{C}$ is naturally isomorphic to $\mathbb{R}^2$. That is, we can define the following map $\phi: \mathbb{C} \to \mathbb{R}^2$:

$$ \phi(a + bi) = \left(\begin{array}{cc} a \\ b \end{array}\right) $$

which is clearly invertible and therefore a bijection between the two. However, it seems weird that $\mathbb{C} \cong \mathbb{R}^2$ just based on intuition. Can anyone give me a better way to think about this isomorphism?

Answer 1

We build the complex numbers $\mathbb C$ by taking the real numbers $\mathbb R$ and adding on the imaginary unit $i$, satisfying $i^2=-1$. But as a vector space, $\mathbb C$ has no multiplicative structure; after all, the vector space axioms say nothing about multiplying vectors, only about adding them. It no longer matters that $i^2=-1$ because we don't know how to multiply two vectors in $\mathbb C$. In that sense $i$ is an arbitrary placeholder -- we could just as well call it $x$, for example, and talk about the vector space of polynomials of the form $a+bx$.

Answer 2

Whilst in some sense you are right, the intuition is not correct.

I mean, it is correct if you think of them both as two real vector spaces; $$\mathbb{R}^2 = span\left\{\left(\begin{matrix} 1 \\ 0\end{matrix}\right),\left(\begin{matrix} 0 \\ 1\end{matrix}\right) \right\},\quad C =span\left\{(1+0i),(0+1i) \right\} $$ then there is indeed an isomorphism between $\mathbb{R}^2$ and $C$ as you described.

The key difference is that $\mathbb{C}$ is a (algebraic) field, so we think of $\mathbb{C}$ as a set of points ($C$ as described above, and thus isomorphic to the euclidian plane), with two closed operators $(+,\cdot)$ which map $C\times C \rightarrow C$.

Now the intuition holds up when we think about the addition operator. Adding vectors in $\mathbb{R}^2$ is equivalent to adding complex numbers. We also have in both, an additive identity and additive inverses.

The key difference is multiplication. A vector space doesn't come with a closed multiplication operator. For instance, the scalar product, which is what comes to mind when we think about multiplication in $\mathbb{R}^2$ doesn't even map back into $\mathbb{R}^2$!

For $\mathbb{C}$ however, if we multiply two complex numbers, we will get a complex number out the other end. I.e. it's closed. (and also has identities, inverses, commutativity and distributive properties.)

In short, $\mathbb{C}$ is a field and as such has a whole bunch more structure than $\mathbb{R}^2$. The intiution is fine as long as you're only thinking about the points, and not operations on those points.