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 $dimV_\mathbb{C} = m$


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$$\dim V_{\mathbb R}=m=\dim V_{\mathbb C}.$$


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