Is $\mathbb{R}^4$ isomorphic to $\mathbb{C}^2$, $\mathbb{R}^4$ isomorphic to $\mathrm{Mat}_{4 \times 4}(\mathbb{R})$ as $\mathbb{R}$-vector spaces?

Question

• Is $\mathbb{R}^4$ isomorphic to $\mathbb{C}^2$ as $\mathbb{R}$-vector spaces?
• Is $\mathbb{R}^4$ isomorphic to $\mathrm{Mat}_{4 \times 4}(\mathbb{R})$ as $\mathbb{R}$-vector spaces?

For the first one, I can define an isomorphism : $\phi:\mathbb{R}^4→\mathbb{C}^2$ as follows:

$$\phi(a,b,c,d)=(a+bi , c+di)$$

This map is linear and bijective, so it is an isomorphism. Is there any better way to think about this as an isomorphism?

The second statement:

$\mathbb{R}^4$ is a 4-dimensional vector space, the $\mathrm{Mat}_{4 \times 4}(\mathbb{R})$ space, consisting of $4 \times 4$ matrices with real entries. Should the dimensionality of $\mathrm{Mat}_{4 \times 4}(\mathbb{R})$ be regarded as $4$, analogous to $\mathbb{R}^4$, or as $16$, considering the total number of real entries in the matrices?

I'm unsure if this is isomorphic and how I can show it.

Answer 1

Comparing $\mathbb R$-dimensions the results follow, since each finite-dimensional $\mathbb R$-vector space $V$ is isomorphic to $\mathbb R^{\text{dim}_{\mathbb R}(V)}$.

Answer 2

Your answer for $\mathbb{R}^4$ and $\mathbb{C}^2$ is correct, they are isomorphic.

But note $\mathrm{Mat}_{4 \times 4}(\mathbb{R})$ is $16$-dimensional. Why is that? Since they are $4 \times 4$ matrices, there are $16$ entries in the matrix, which all correspond to a dimension of the vector space if we choose a certain basis - namely the matrices who are zero in all but one entry.