Is the finite dimension of a vector space over the complex numbers half the dimension of the same vector space considered over the reals?

Question

Consider a vector space V with basis ${b_{1},..,b_{n}}$ and complex scalars. This obviously has dimension n. Now consider a space with the same exact set of vectors of V, except with real scalars.

I am thinking the basis of this space would be ${b_{1},...b_{n}, ib_{1}, ...ib_{n}}$ which would have dimension 2n.

Is this correct?

The proof is as follows: $b_{k}$ and $ib_{k}$ are obviously elements of the space. With the complex scalars, $b_{k}$ and $ib_{k}$ are dependent.

When considering the same space with real scalars $b_{k}$ and $ib_{k}$ must be independent. Also, {i$b_{k}$, $b_{1}$, $b_{n}$} must be independent, since otherwise considered as a complex space, {$b_{1}$,..$b_{n}$} would be dependent, which as a basis cannot be.

Is this conclusion and proof correct?

Answer 1

Let $V$ be the space in question.

Then $b_1,...,b_n$ is a basis for $V$ over $\mathbb{C}$ iff $b_1,...,b_n, i b_1,...,i b_n$ is a basis for $V$ over $\mathbb{R}$.

($\Rightarrow$): If $\sum_k \alpha_k b_k + \sum_k \beta_k (i b_k) = 0$ (with $\alpha_k , \beta_k \in \mathbb{R}$) then $\sum_k (\alpha_k+i \beta_k) b_k = 0$ and hence $\alpha_k=\beta_k = 0$. If $v \in V$, then $v = \sum_k \gamma_k b_k$, where $\gamma_k \in \mathbb{C}$. Hence $v= \sum_k (\text{re } \gamma_k) b_k + \sum_k (\text{im } \gamma_k) ib_k$, and so $b_1,...,b_n, i b_1,...,i b_n$ is a basis for $V$ over $\mathbb{R}$.

($\Leftarrow$): Suppose $\sum_k \gamma_k b_k =0$ where $\gamma_k \in \mathbb{C}$. Then $\sum_k (\text{re } \gamma_k) b_k + \sum_k (\text{im } \gamma_k) ib_k = 0$ and so it follows that $\text{re } \gamma_k = \text{im } \gamma_k = 0$ and so $\gamma_k = 0$. If $v \in V$ then $v=\sum_k \alpha_k b_k + \sum_k \beta_k (i b_k) = 0$ where $\alpha_k , \beta_k \in \mathbb{R}$. Then $v = \sum (\alpha_k + i \beta_k) b_k$ and hence $b_1,...,b_n$ is a basis for $V$ over $\mathbb{C}$.

It follows that a space has $\dim$ n over $\mathbb C$ iff the space has $\dim$ 2n over $\mathbb R$.

Answer 2

I think, the answer is that the question is flawed: with real scalar, it won't be "the same space." For example, suppose, with complex scalars, we have a 2-D space, ${\bf C}^{2}$, with basis $$ b_{1} = (2 + i3, 0), \; b_{2} = (0, 3 - 2i). $$ What would be "the same vector space with real scalars"?

Now, if you were thinking of complexification or realification, that's different, and will have a different formulation, depending on what you wanted to ask.