Exercise 9.A.1

Proof. $V_{ℂ}$ is clearly closed under addition. We can write each complex number in the form $a+bi$ for some $a,b\in ℝ$ and we have

where the first equality follows from the definition and the second because the vectors inside both parentheses are in $V$. Thus $V_{ℂ}$ is closed under complex scalar multiplication. If $0$ is the additive identity on $V$, then $0+i0$ is the additive identity on $V_{}$

C$because$

One easily checks that the rest of the properties listed in 1.19 are satisfied by $V_{ℂ}$.

Therefore $V_{ℂ}$ is a complex vector space. □