How to prove that complex numbers $(\mathbb{C})$ are linear space over real numbers $(\mathbb{R})$ field? How to prove that complex numbers $\mathbb{C}$ are linear space over real numbers field $\mathbb{R}$ ?
 A: $\mathbb{C} = \mathbb{R}(i) = \{a+ib:a,b \in \mathbb{R} \}$ ($\mathbb{R}(i)$ is the smallest field which contains $\mathbb{R}$ and $i$). You can see that it is a 2-dimention vector space on $\mathbb{R}$ ( for example, you can verify that $\{1,i\}$ is a basis for it).
I hope I've helped you.
A: Verify that it satisfies the axioms, of course!


*

*The sum of two complex numbers is a complex number.

*Multiplying a complex number by a real number yields a complex number.

*Addition of complex numbers is associative.

*Addition of complex numbers is commutative.

*The complex number $0$ is such that $z+0 = z$ for all complex numbers $z$.

*Given a complex number $z$, there is a complex number (namely, $-z$) such that $z+(-z) = 0$.

*Multiplication by reals numbers is associative: $\alpha(\beta z) = (\alpha\beta)z$ for all $\alpha,\beta\in\mathbb{R}$, $z\in \mathbb{C}$.

*Multiplication by reals distributes over the sum of complex numbers: $\alpha(z+w) = \alpha z + \alpha w$ for all $\alpha\in\mathbb{R}$, $z,w\in\mathbb{C}$.

*Multiplication by a complex number distributes over the sum of real numbers: $(\alpha+\beta)z = \alpha z + \beta z$ for all $\alpha,\beta\in\mathbb{R}$, $z\in\mathbb{C}$.

*Multiplication by the real number $1$ is the identity mapping: $1z = z$ for all $z\in\mathbb{C}$.


If all 10 are true, then $\mathbb{C}$ is a vector space/linear space over $\mathbb{R}$.
(More generally, if $\mathbf{F}$ is a field, and $\mathbf{K}$ is a field that contains $\mathbf{F}$, then $\mathbf{K}$ is a vector space over $\mathbf{F}$). 
A: The term "linear space" is another word for vector space. There are two aspects to showing that something (call it $V$) is a vector space over a field $K$ (in this case, $K=\mathbb{R}$, the real numbers): 
First, you need to specify a vector addition operation "$\star$", that takes any two elements of $V$ and produces another one, and a scalar multiplication operation "$\circ$", that takes an element of $K$ and an element of $V$ and produces an element of $V$.  Second, you need to show that those operations satisfy the axioms for being a vector space.
Here, the set $V$ under consideration is $\mathbb{C}$, the complex numbers. The vector addition is that is implied is just the usual addition of complex numbers; that is, we define
$$a\star b:=a+b\text{ for all }a,b\in\mathbb{C}.$$
The scalar multiplication (i.e., multiplying a vector (an element of $V$) by a scalar (an element of $K$)) that is implied is just the multiplication of a complex number by a real number; that is, we define
$$\lambda\circ a:=\lambda a\text{ for all }\lambda\in\mathbb{R}\text{ and }a\in\mathbb{C}.$$
The axioms for being a vector space are here. There is not really anything substantial you need to show. For example, $1\circ a=a$ for any $a\in\mathbb{C}$ because, by definition,
$$1\circ a=1a=a$$
(because $1\in\mathbb{R}\subset\mathbb{C}$ is the multiplicative identity of $\mathbb{C}$'s usual multiplication). Another example:
$$\lambda\circ (a\star b)=\lambda(a+b)=\lambda a +\lambda b=(\lambda\circ a)\star(\lambda\circ b)$$
because multiplication of complex numbers distributes over addition.
A: $\mathbb{C}=\{a+b\mathrm{i}\mid a,b\space \mathrm{from}\space \mathbb{R}\}$  and $u, v$ vectors from $\mathbb{C}$. $u=(a_1+b_1\mathrm{i})$ and $v=(a_2+b_2\mathrm{i})$.
To show that $\mathbb{C}$ is a linear space over $\mathbb{R}$, we can prove it by this:
$u+v=((a_1+b_1\mathrm{i})+(a_2+b_2\mathrm{i}))=(a_1+a_2)+(b_1\mathrm{i}+b_2\mathrm{i})=(a_1+a_2)+(b_1+b_2)i$ so the new vector formed by summary of two other vectors has the form $A+B\mathrm{i}$.
This means that the new vector is part of $\mathbb{C}$, and by this we prove that $\mathbb{C}$ is a linar space over $\mathbb{R}$ field.
