The complex line through (1,i) has a two-vector real basis

The coordinate sum norm on $V$ is a complex norm by [F2] and [F1]. The set $W$ contains zero because $0_{\mathbb{C}}b=0_V$ by [F15]. The vector-space laws [F12] give $\lambda b+\mu b=(\lambda +\mu )b$ and $\alpha (\lambda b)=(\alpha \lambda )b$, so $W$ is closed under addition and complex scalar multiplication; [F7] makes it a complex-linear subspace. The restricted norm inherits separation, full complex homogeneity, and the triangle inequality, so it gives the complex normed subspace in [F3].

If $r(1,i)+s(i,-1)=(0,0)$ for $r,s\in \mathbb{R}$, the pointwise coordinate operations of [F14] give first coordinate $r+is=0$. Uniqueness of the real and imaginary coordinates in [F6] implies $r=s=0$, so the list is linearly independent by [F9].

Applying [F4] to this complex normed subspace gives the real normed subspace $W_{\mathbb{R}}\subseteq V_{\mathbb{R}}$ and the identical restricted norm metric.

For $\lambda =a+ic$, [F6] gives $\lambda b=(a+ic,i(a+ic))=(a+ic,-c+ia)=a(1,i)+c(i,-1)$ using coordinatewise scalar multiplication [F14] and the vector-space laws [F12]. Thus every element of $W_{\mathbb{R}}$ is a real linear combination of the displayed list. Conversely, $b\in W$ by taking $\lambda =1$ and $ib\in W$ by taking $\lambda =i$; since $W$ is a complex-linear subspace by step 1.1 and real scalars act through the canonical embedding [F5], every real linear combination of $b$ and $ib$ belongs to $W_{\mathbb{R}}$. The two inclusions and [F10]–[F11] show that the list spans $W_{\mathbb{R}}$.

By steps 2.2 and 1.2 the list is spanning and linearly independent. By [F13] it is injective with linearly independent image, and step 2.2 says that image spans; hence [F8] makes it an ordered real basis.