$\mathbb{R}^2$ is not a subspace of $\mathbb{C}^2$?

I found a answer to a question with an answer which shocked me a little, so I wanted to make sure it is correct.

The answer told, that if $\mathbb{R}^2$ is a subspace of $\mathbb{C}^2$, then $$i(1,1)=(i,i)$$ belongs to $\mathbb{R}^2$ which is a contradiction.

I agree that this is a contradiction, but being a subspace in this case mean that there is a $(0,0)$ element and the set is closed under addition and scalar multiplication. The $\mathbb{C}^2$ includes all the elements of $\mathbb{R}^2$, so why it cannot be a subspace?

Is there a condition that space and subspace must be defined over the same field?

It is from Sheldon Axler "Linear Algebra Done Right" pg. 24, ex. 5.

Answer found here https://linearalgebras.com/1c.html.

Thanks a lot!

Solution 1:

• $\mathbb{R}^{2}$ is a $\mathbb{R}$-vector-space of dimension 2.
• $\mathbb{C}^{2}$ is a $\mathbb{C}$-vector-space of dimension 2, and an $\mathbb{R}$-vector-space of dimension 4.

As an $\mathbb{R}$-vector-space, $\mathbb{R}^2 < \mathbb{C}^2$. As a $\mathbb{C}$-vector-space, $\mathbb{R}^2 \not< \mathbb{C}^2$.