(related to Proposition: Complex Numbers as a Vector Space Over the Field of Real Numbers)
We will prove that the complex numbers \(\mathbb C\) form a vector space over the field \(\mathbb R\) with respect to the following maps: \[ \cases{ \mathbb C\times \mathbb C\mapsto \mathbb C:x,y\mapsto x + y:=(a+c,b+d) & \text{(the vector addition)}\\ \mathbb R\times \mathbb C\mapsto \mathbb C:\alpha,x\mapsto \alpha \cdot x:=(\alpha \cdot a,\alpha \cdot b) & \text{(the scalar multiplication)} }\] for any real number \(\alpha\in\mathbb R\) and any complex numbers \(x,y\in\mathbb C\), identified by some ordered pairs of real numbers \(x:=(a,b)\), \(y:=(c,d)\), \(a,b,c,d\in\mathbb R\).
In particular, we have shown that the real numbers together with addition and multiplication form a field (\(\mathbb R,+,\cdot)\). We have also shown that the complex numbers together with the addition form a commutative group \((\mathbb C, + )\), in particular they form a non-empty set of vectors.
It remains to be shown that for the real numbers \(\alpha,\beta\in\mathbb R\) and the complex numbers \(x,y\in\mathbb C\), the following rules of scalar multiplication hold: 1. \((\alpha + \beta)\cdot x=\alpha\cdot x + \beta\cdot x\), 1. \(\alpha\cdot(x + y)=\alpha\cdot x + \alpha\cdot y\), 1. \((\alpha\cdot \beta)\cdot x=\alpha\cdot (\beta\cdot x)\), 1. If \(1\) is the neutral element of multiplication in \(\mathbb R\), then it is also a neutral element of the scalar multiplication in \(\mathbb C\), i.e. \(1\cdot x=x\).
Using the definition of complex numbers, the above definition of scalar multiplication, the distributivity rule for real numbers and the definition of adding complex numbers, we can confirm the first rule of scalar multiplication:
\[\begin{array}{rcll} (\alpha + \beta)\cdot x&=&(\alpha + \beta)\cdot (a,b)&\text{by definition of complex numbers}\\ &=&((\alpha + \beta)\cdot a,(\alpha + \beta)\cdot b)&\text{by definition of scalar multiplication }\\ &=&((\alpha\cdot a) + (\beta\cdot a),(\alpha\cdot b) + (\beta\cdot b))&\text{by distributivity rule for real numbers}\\ &=&(\alpha\cdot a,\alpha\cdot b) + (\beta\cdot a, \beta\cdot b)&\text{by definition of adding complex numbers}\\ &=&\alpha\cdot x + \beta\cdot x &\text{by definition of complex numbers} \end{array}\]
In analogy to 1., we can prove the second rule of scalar multiplication:
\[\begin{array}{rcll} \alpha\cdot(x + y)&=&\alpha\cdot((a,b)+ (c,d))&\text{by definition of complex numbers}\\ &=&\alpha\cdot(a+c,b+d)&\text{by definition of adding complex numbers}\\ &=&(\alpha\cdot(a+c),\alpha\cdot(b+d))&\text{by definition of scalar multiplication}\\ &=&((\alpha\cdot a) + (\alpha\cdot c),(\alpha\cdot b) + (\alpha\cdot d))&\text{by distributivity rule for real numbers}\\ &=&(\alpha\cdot a,\alpha\cdot b) + (\alpha\cdot c, \alpha\cdot d)&\text{by definition of adding complex numbers}\\ &=&\alpha\cdot (a,b) + \alpha\cdot (c,d)&\text{by definition of scalar multiplication}\\ &=&\alpha\cdot x + \alpha\cdot y &\text{by definition of complex numbers} \end{array}\]
Using the multiplying real numbers is associative, we can confirm the third rule of scalar multiplication:
\[\begin{array}{rcll} (\alpha \cdot \beta)\cdot x&=&(\alpha \cdot \beta)\cdot (a,b)&\text{by definition of complex numbers}\\ &=&((\alpha \cdot \beta)\cdot a,(\alpha \cdot \beta)\cdot b)&\text{by definition of scalar multiplication }\\ &=&(\alpha \cdot (\beta\cdot a),\alpha \cdot (\beta\cdot b))&\text{by associativity of multiplying real numbers}\\ &=&\alpha \cdot (\beta\cdot a,\beta\cdot b)&\text{by definition of scalar multiplication}\\ &=&\alpha \cdot (\beta\cdot (a,b))&\text{by definition of scalar multiplication}\\ &=&\alpha\cdot (\beta\cdot x)&\text{by definition of complex numbers} \end{array}\]
We have shown the existence of \(1\) as a neutral element of multiplication of real numbers. Obviously,
\[\begin{array}{rcll} 1\cdot x&=&1\cdot (a,b)&\text{by definition of complex numbers}\\ &=&(1\cdot a,1\cdot b)&\text{by definition of scalar multiplication }\\ &=&(a,b)&1\text{ is neutral with respect to the multiplication of real numbers}\\ &=&x&\text{by definition of complex numbers} \end{array}\]
