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how to prove that complex numbers $mathbbC$ are linear space over real numbers field $mathbbR$ ? it's a linear algebra question

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, $K=mathbbR$, 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 $mathbbC$, the complex numbers. The vector addition is that is implied is just the usual addition of complex numbers; that is, we define
$$astar b:=a+btext for all a,binmathbbC.$$
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
$$lambdacirc a:=lambda atext for all lambdainmathbbRtext and ainmathbbC.$$

The axioms for being a vector space are here. There is not really anything substantial you need to show. For example, $1circ a=a$ for any $ainmathbbC$ because, by definition,
$$1circ a=1a=a$$
(because $1inmathbbRsubsetmathbbC$ is the multiplicative identity of $mathbbC$'s usual multiplication). Another example:
$$lambdacirc (astar b)=lambda(a+b)=lambda a +lambda b=(lambdacirc a)star(lambdacirc b)$$
because multiplication of complex numbers distributes over addition.

Verify that it satisfies the axioms, of course!

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


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


3. Addition of complex numbers is associative.


4. Addition of complex numbers is commutative.


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


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


7. Multiplication by reals numbers is associative: $alpha(beta z) = (alphabeta)z$ for all $alpha,betainmathbbR$, $zin mathbbC$.


8. Multiplication by reals distributes over the sum of complex numbers: $alpha(z+w) = alpha z + alpha w$ for all $alphainmathbbR$, $z,winmathbbC$.


9. Multiplication by a complex number distributes over the sum of real numbers: $(alpha+beta)z = alpha z + beta z$ for all $alpha,betainmathbbR$, $zinmathbbC$.


10. Multiplication by the real number $1$ is the identity mapping: $1z = z$ for all $zinmathbbC$.

If all 10 are true, then $mathbbC$ is a vector space/linear space over $mathbbR$.

(More generally, if $mathbfF$ is a field, and $mathbfK$ is a field that contains $mathbfF$, then $mathbfK$ is a vector space over $mathbfF$).

$mathbbC = mathbbR(i) = a+ib:a,b in mathbbR $ ($mathbbR(i)$ is the smallest field which contains $mathbbR$ and $i$). You can see that it is a 2-dimention vector space on $mathbbR$ ( for example, you can verify that $1,i$ is a basis for it).

I hope I've helped you.

$mathbbC=a+bmathrmimid a,bspace mathrmfromspace mathbbR$ and $u, v$ vectors from $mathbbC$. $u=(a_1+b_1mathrmi)$ and $v=(a_2+b_2mathrmi)$.
To show that $mathbbC$ is a linear space over $mathbbR$, we can prove it by this:
$u+v=((a_1+b_1mathrmi)+(a_2+b_2mathrmi))=(a_1+a_2)+(b_1mathrmi+b_2mathrmi)=(a_1+a_2)+(b_1+b_2)i$ so the new vector formed by summary of two other vectors has the form $A+Bmathrmi$.
This means that the new vector is part of $mathbbC$, and by this we prove that $mathbbC$ is a linar space over $mathbbR$ field.