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 $\mathbb{R}$ ?
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 $\mathbf{F}$).
$\mathbb{C}=\{a+b\mathrm{i}\mid a,b\space \mathrm{from}\space \mathbb{R}\}$  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.