(related to Lemma: Linear Independence of the Imaginary Unit \(i\) and the Complex Number \(1\))
Because we have already shown that the complex numbers form a vector space over the field of real numbers, we will now prove that the vectors complex number one \(1\) and the imaginary unit \(i\) are linearly independent. Let \(\alpha,\beta\in\mathbb R\) be some real numbers and consider the equation \[\alpha\cdot 1 + \beta \cdot i=0.\] We have to prove that this equation has only the trivial solution \(\alpha=\beta=0\). For the purpose of a better understanding of the prove, we will use the notation \[\alpha\cdot (1) + \beta \cdot (i)=0,\] to indicate that \((1)\) and \((i)\) are vectors and not simply numbers.
By definition, the complex vector \(1:=(1,0)\), i.e. the complex number \(1\) is identical with the ordered pair of real numbers \((1,0)\). Analogously, the complex vector \(i:=(0,1)\), i.e. the complex number \(i\) is identical with the ordered pair of real numbers \((0,1)\). Therefore, the above equation is equivalent to \[\begin{array}{rcl} \alpha\cdot (1) + \beta \cdot (i)=0&\Longleftrightarrow&\beta\cdot (i)=-\alpha\cdot (1)\\ &\Longleftrightarrow&\beta\cdot (0,1)=-\alpha\cdot (1,0) \end{array}\] From the definition of the scalar multiplication, it follows \[ (0,\beta)=(-\alpha,0)\Longleftrightarrow -\alpha=0\text{ and }\beta=0.\]
Because \(-\alpha=-1\cdot \alpha\) (i.e the product of the real numbers \(-1\) and \(\alpha\)) and because the product of two real numbers is only zero, if at least one of them is zero, it follows \(\alpha=0\).
