How is $\mathbb{C}\times\mathbb{C}$ a real vector space? I'm working on Linear Algebra homework. I'm having trouble with: $\mathbb{C}\times\mathbb{C}$ is a real vector space. Explain why. Write down a basis for this real vector space.
I'm just confused on what exactly $\mathbb{C}\times\mathbb{C}$ is and why it is a real vector space. It seems it can be defined for vector addition and scalar multiplication. I've seen $+\colon\mathbb{C}\times\mathbb{C}$ and $*\colon\mathbb{C}\times\mathbb{C}$ and I'm confused because I thought it was a cross product?
I know once I figure out how to define $\mathbb{C}\times\mathbb{C}$, I have to show that the vector space properties apply.
Thanks in advance for any help.
 A: A vector space is defined as a quadruple $(\mathbf{V},\mathbb{K},\oplus,\odot)$ where $\mathbf{V}$ is a set of elements called vectors, $\mathbb{K}$ is a field $(\mathbb{K},+,\cdot)$ , $\oplus$ is a binary operation (called sum) on $\mathbf{V}$ such that $(\mathbf{V},\oplus)$ is a commutative group and $a\odot\mathbf{v}:\mathbb{K}\times\mathbf{V} \rightarrow \mathbf{V}$ is a scalar multiplication such that, $\forall a,b \in \mathbb{K}$ and $\forall \mathbf{u,v} \in \mathbf{V}$ we have:
$$
a\odot(b\star\mathbf{v})=(a\cdot b)\odot\mathbf{v}  
$$
$$
1\odot\mathbf{v}=\mathbf{v}
$$
$$
a \odot (\mathbf{u}\oplus\mathbf{v})=a \odot\mathbf{u}\oplus a\odot \mathbf{v}
$$
$$
(a+b)\odot \mathbf{v}=a\odot \mathbf{v}\oplus b\odot \mathbf{v}
$$
Note that $(+,\cdot)$ are the operations on $\mathbb{K}$ and are different from the operations $(\oplus, \odot)$.
In your case $\mathbf{V}= \mathbb{C}\times \mathbb{C}$ i.e the vectors are couple $(\alpha_1,\alpha_2)=\mathbf{a}$ with $\alpha_1,\alpha_2 \in \mathbb{C}$.
Usually we define the operations $\oplus$ as:
$$
\mathbf{a}\oplus\mathbf{b}=(\alpha_1,\alpha_2)\oplus(\beta_1,\beta_2)=(\alpha_1+\beta_1,\alpha_2+\beta_2)
$$
Where the $+$ operation is the usual sum in $\mathbb{C}$ and, since $(\mathbb{C},+)$ is a commutative group, we can easily see that $(\mathbb{C}\times\mathbb{C},\oplus)$ is a commutative group, with neutral element $(0,0)$ and opposite of $(\alpha_1,\alpha_2)$ the element $(-\alpha_1,-\alpha_2)$.
Now we want $\mathbb{K}=\mathbb{R}$ and we define the scalar multiplication as:
$$
r\odot \mathbf{a}=r\odot(\alpha_1,\alpha_2)=(r\cdot\alpha_1,r\cdot \alpha_2) 
$$
Where $ r\cdot \alpha_i$ is the product in $\mathbb{C}$ and is well defined also for $r\in \mathbb{R}$ and $\alpha_i \in \mathbb{C}$ since $\mathbb{R}$ is a subfield of $\mathbb{C}$.
Now, using the properties of $(\mathbb{C},+,\cdot)$ as a field it is not difficult to see that all the axioms for a scalar multiplication are verified. 
As an example we have:
$$
r\odot(\mathbf{a}\oplus\mathbf{b})=r\odot\left( (\alpha_1,\alpha_2)\oplus(\beta_1,\beta_2)\right)=
r\odot\left( (\alpha_1+\beta_1),(\alpha_2+\beta_2)\right)=
$$
$$
= \left( r\cdot(\alpha_1+\beta_1),r\cdot(\alpha_2+\beta_2)\right) =
\left( r\cdot\alpha_1+r\cdot\beta_1,r\cdot\alpha_2+r\cdot\beta_2 \right)=
$$
$$
=r\odot(\alpha_1,\alpha_2)\oplus r\odot(\beta_1,\beta_2)= r\odot\mathbf{a}\oplus r\odot\mathbf{b}
$$ 
