Is $\mathbb{C}^2$ isomorphic to $\mathbb{R}^4$?

Are $\mathbb{C}^2$ and $\mathbb{R}^4$ isomorphic to one another? Two vector spaces are isomorphic if and only if there exists a bijection between the two. We can define the linear map $T: \mathbb{C}^2 \mapsto \mathbb{R}^4$ as

T\left(\left[\begin{array}{cc} a + bi \\ c + di \end{array}\right]\right) = \left[\begin{array}{cccc} a \\ b \\ c \\ d \end{array}\right]

Does this bijection suffice to show that they’re isomorphic?

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As T. Bongers points out, we also need the map $T$ to be linear if we would like it to be an isomorphism. It isn’t too hard to see that $T(v + w) = T(v) + T(w)$, and if $\alpha \in \mathbb{R}$, then $T(\alpha v) = \alpha T(v)$. This shows that as real vector spaces, $\mathbb{C}^2$ and $\mathbb{R}^4$ are isomorphic. However, for $\alpha \in \mathbb{C}$, we do not have $T(\alpha v) = \alpha T(v)$ because the right hand side isn’t defined, this is because $\mathbb{R}^4$ is not a complex vector space.

In summary, $\mathbb{C}^2$ and $\mathbb{R}^4$ are isomorphic as real vector spaces, but $\mathbb{C}^2$ is also a complex vector space, while $\mathbb{R}^4$ is not.

Saying things a bit differently than the other answers: Yes, what you have is an vector space isomorphism. When we talk about $\mathbb{R}^4$ we (usually) don’t consider it as a complex vector space, so it is fairly safe to assume that both your spaces are real vector spaces and as such they are isomorphic (as mentioned in the other answers as well).

So is $\mathbb{R}^4$ really not a complex vector space? Well, using the bijection $T$ that you have, you can actually make $\mathbb{R}^4$ into a complex vector space by, for $\lambda\in \mathbb{C}$ and $a\in \mathbb{R}^4$, defining the scalar multiplication:
\lambda.a = T(\lambda T^{-1}(a)).
With this definition, you then have an isomorphism of complex vector spaces. So just for fun you would get
i\pmatrix{1 \\ 0 \\ 0\\ 0} = T\pmatrix{i \\ 0} = \pmatrix{0 \\ 1 \\ 0 \\ 0}
And here are two simple bonus questions: What is the dimension of $\mathbb{R}^4$ view as a complex vector space this way? What is a basis?

If you are talking about isomorphism as vector spaces over $\mathbb{R}$ then yes, you are correct. However a bijection between two spaces is not enough for them to be isomorphic ( for example $\mathbb{R}$ is bijectable with $\mathbb{C}$ but they are not isomorphic as vector spaces over $\mathbb{R}$. What you need is a bijection that is also a linear function

Your map is indeed an isomorphism, but I want to make another point, as well.

You’re almost right about bijections, but a bijection is not enough to show that they are isomorphic (an easy example is given by AngelTC). However, if you have a bijection between the bases of these two spaces as $\mathbb{R}$-vector spaces, then you can conclude that they are isomorphic, since a bijection between bases can be extended linearly to create the isomorphism.

In general vector spaces, if you have spaces $V$ and $V’$ with bases $\left\{b_{1},\ldots,b_{n}\right\}$ and $\left\{b’_{1},\ldots,b’_{n}\right\}$, respectively, you can form the bijection $f\left(b_{i}\right)=b’_{i}$, and then extend it “linearly” by just saying that $f\left(b_{i}+b_{j}\right)=b’_{i}+b’_{j}$ and $f\left(\alpha b_{i}\right)=\alpha b’_{i}$ where $\alpha$ is a scalar. This turns a bijection between bases into an isomorphism.

So, take your favorite basis for $\mathbb{R}^{4}$ and your favorite basis for $\mathbb{C}^{2}$. Can you form a bijection between these? If so, then just extend linearly for an isomorphism.