Is $\Bbb R^2$ a subspace of the complex vector space $\Bbb C^2$

Question

I am attempting to illustrate whether or not $\mathbb{R}^2$ is a subspace of the complex vector space $\mathbb{C}^2$. Originally my intuition said yes, because we can write any real number as a complex number where $b=0$, ie we can write $2 \in \mathbb{R}$ as $2=2+0i$.

However, this is apparently incorrect, and the example in my book shows that this would suggest that $i(1,1) = (i,i) \in \mathbb{R}^2$ and thereby yielding a contradiction.

I guess I don't understand the above argument. Is it because $\mathbb{R}^2$ is not closed under multiplication when we consider a scalar $a \in \mathbb{F}^2,$ where $\mathbb{F}$ represents a field of $a \in \mathbb{R}^2$?

Any help would be appreciated.

Answer 1

It depends on your field. If $\Bbb F=\Bbb R$ then $\Bbb R^2$ is indeed a subspace. If $\Bbb F=\Bbb C$ then $\Bbb R^2$ is NOT a subspace.

Answer 2

To say that $\mathbb{R}^2$ is a $\mathbb{C}$-subspace of the $\mathbb{C}$-vector space $\mathbb{C}^2$ implies that \begin{equation} \forall \alpha \in \mathbb{C}, \forall x \in \mathbb{R}^2, \quad \alpha \cdot x \in \mathbb{R}^2 \end{equation} since a vector space must contain the result of any of its vectors multiplied by any of the scalars.

With $\alpha = i$ and $x = (1, 1)$ we find a vector $\alpha \cdot x = (i, i)$ which is not an element of $\mathbb{R}^2$.