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