Let \(V\) be the set of complex numbers and let \(F\) be the field of real numbers. With the usual operations, \(V\) is a vector space over \(F\). Describe explicitly an isomorphism of this space onto \(\RR^2\).
Solution
Define \[ \Phi:\mathbb C\longrightarrow \mathbb R^2,\qquad \Phi(x+iy)=(x,y). \] For real scalars \(a,b\) and complex numbers \(z,w\), \[ \Phi(az+bw)=a\Phi(z)+b\Phi(w), \] so \(\Phi\) is real-linear. It is injective because \(\Phi(x+iy)=(0,0)\) implies \(x=y=0\), and it is surjective because \((x,y)=\Phi(x+iy)\). Hence \(\Phi\) is an isomorphism of real vector spaces.