# Dimension of a Subspace?

Let $$V_\mathbb{R}$$ be a subspace of $$\mathbb{R}^n$$ over the field of real numbers. Let $$\{v_1, \cdots, v_m\}$$ be a basis of $$V_\mathbb{R}$$, i.e., $$x \in V_\mathbb{R} \Longleftrightarrow x = \sum_{i=1}^m \, a_iv_i, a \in \mathbb{R}$$. Let the complexification $$V_\mathbb{C}$$ of $$V_\mathbb{R}$$ be the subspace of $$\mathbb{C}^n$$ over the field of complex numbers defined as follows: $$V_\mathbb{C} = \{z \in \mathbb{C}^n \, : \, z = x + jy, \, x \in V_\mathbb{R}, \, y \in V_\mathbb{R} \}$$ where $$j := \sqrt-1$$.

Prove that $$dimV_\mathbb{C} = m$$

Question:

I am slightly confused about what is being asked above. Since the basis of $$V_\mathbb{R}$$ is $$\{v_1, \cdots, v_m \}$$ wouldn't the basis of $$V_\mathbb{C}$$ also have a dimension of $$m$$?

I am assuming this since $$V_\mathbb{C}$$ is simply a combination of $$x$$ and $$y$$ elements contained in $$V_\mathbb{R}$$

Is the question basically asking me to prove $$V_\mathbb{C}$$ is closed under addition and scalar multiplication? which would imply $$V_\mathbb{C}$$ has the same dimension as the basis of $$V_\mathbb{R}$$?

First of all, in general there is no such thing as as the basis of a vector space, since vector spaces (over $$\mathbb R$$) have infinitely many bases (again, in general).
If $$\{v_1,\ldots,v_m\}$$ is a basis of $$V_{\mathbb R}$$, then I suggest that you prove that $$\{v_1,\ldots,v_m\}$$ is also a basis of $$V_{\mathbb C}$$. After that, you can state that$$\dim V_{\mathbb R}=m=\dim V_{\mathbb C}.$$