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To find is a $\mathbb{C}$ vector space structure over $\mathbb{R}$ with addition and scalar multiplication such that scalar multiplication has the following property:

$\cdot : \mathbb{C} \times \mathbb{R} \rightarrow \mathbb{R}$ s.t.

$\cdot : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is a usual multiplication in $\mathbb{R}$

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If you don't require any compatibility on the pre-existing $\mathbb{R}$-vector space structure on $\mathbb{R}$, then there is a $\mathbb{C}$-vector space structure on the set $\mathbb{R}$ simply because $\mathbb{R}$ and $\mathbb{C}$ have the same cardinality. –  Zhen Lin Nov 23 '12 at 11:16
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No. For any vector space $V$, any nonzero vector $v$, and any scalars $k$ and $l$, $kv=lv$ implies that $k=l$. Here, $kv$ already takes all possible values when $k$ ranges over $\Bbb R$, so there's nothing for (say) $iv$ to be.

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Let $V$ be a real vector space. Using axioms of vector spaces you can prove that you only need to find out what $\imath \cdot v$ should be for $v\in V$. So let $$ \phi(x) = \imath \cdot x $$

Since $r_1 \cdot (r_2 \cdot v) = r_1r_2 \cdot v$ you get for $r_1 = r_2 = \imath$ the following equality must hold $$ \phi(\phi(v)) = -v $$ for all $v \in V$. Moreover, again using vector space axioms you can prove that $$ \phi(a\cdot x + b \cdot y) = a\phi(x) + b\phi(y) $$ for all $a,b \in \mathbb{R}$ and all vectors $x,y \in V$.

So existence of such a linear mapping $\phi: V \to V$ that squares to minus identity is equivalent to possibility of extending your real vector space structure to complex vector space structure.

The rest is up to you. Hint: Suppose such a $\phi$ exists. What can you say about the dimension of $V$?

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This supposed vector space would be $1-$dimensional. Let $\{a\}$ be a basis. Then $i\cdot a = r =\frac{r}{a}\cdot a , \ \ r \in \mathbb{R}$ ↯.

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$\{a \neq 0\}$ is not a basis ... –  Martin Brandenburg Nov 23 '12 at 9:39
    
@MartinBrandenburg: I didn't say is a basis. If $\mathbb{R}$ was a vector space over $\mathbb{C}$ with the described properties then necessarily $\dim_{\mathbb{C}}\mathbb{R}=1$. Thus in that case $\{a\}$ for $a\neq 0$ would be a basis. –  P.. Nov 23 '12 at 9:52
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