Is the following Proof Correct?
Theorem. $\mathbf{R}^{2}$ is not a subspace of $\mathbf{C}^{2}$.
Proof. Consider the set $\mathbf{C^{'}}$ which we define as follows
$$\mathbf{C^{'}}=\{(x_1,x_2)\in\mathbf{C}^2\mid\Im(x_1)=\Im(x_2)=0\}\tag{1}$$
It is evident that $\mathbf{R^{2}}=\mathbf{C^{'}}$. Now assume that $\mathbf{R}^2$ is a subspace of $\mathbf{C}^{2}$ and consider the following argument.
Let $\lambda=a+bi\in\mathbf{C}$ and $u=(u_1+0i,u_2+0i)\in\mathbf{C^{'}}$ such that $a\neq 0,b\neq0,u_1\neq 0$ and $u_2\neq0$ thus $\lambda u=(au_1+bu_1i,au_2+bu_2i)$ but
$$\Im(au_1+bu_1i)=bu_1\neq 0\tag{2}$$
$$\Im(au_2+bu_2i)=bu_2\neq 0\tag{3}$$
implying that $\lambda u\not\in \mathbf{C^{'}}$,but $\mathbf{C^{'}}$ is a subspace of $\mathbf{C^2}$ and is therefore closed under scalar multiplication, resulting in a contradiction.
$\blacksquare$
Best Answer
It depends on what you consider to be "scalars". If you take "scalar" to mean "real number" then $\mathbb R^2$ is a subspace of $\mathbb C^2,$ since it's closed under linear combinations, which is because the coefficients in linear combinations are real numbers.
But if you take "scalar" to mean "complex number" then it's not a subspace since it's not closed under linear combinations in which some of the scalar coefficients are not real.
Or in other words: You haven't fully specified which vector space you're talking about until you say what the scalars are. The set of scalars has to be closed under addition, subtraction, multiplication, and division, i.e. it needs to be a field, so for example $\mathbb Q$ could be the field of scalars.