TITLE: Extending a real vector space into a complex vector space using a linear map
QUESTION [7 upvotes]: Let $V$ be real $n$-dimensional vector space, and $T:V\to V$ is a linear map satisfying the condition $T^2(v)=-v$ for all $v \in V$. Then,

Show that $n$ is an even integer.
Use $T$ to make $V$ into a complex vector space such that the multiplication by complex numbers extends the multiplication by reals.
Show that, with respect to the complex vector space structure on $V$ obtained in 2, $T:V\to V $ is a complex linear function.


This problem is bugging me for a while. And I have a few questions about it. I did no. 1 using the concept of minimal polynomials. [Another nice proof can be found here.] But the real troubles are question no. 2 and 3. The whole statement of Q.2 looks very vague to me. (For instance, I have doubts that, if I declare a real vector space to be a complex one, how can it be same as the previous one?) The follow up question has an equally dubious statement.  
I would be glad if somebody takes the time to clarify what these two statements actually mean and exactly what I have to prove. Thank you.
[Source: This question can be found here (Question 25b).]

REPLY [3 votes]: Define complex scalar multiplication $ℂ×V→V$ by $(a+ib)v = av + b T(v)$. Prove that this defines a vector space. (The crucial point is that associativity law concerning scalar multiplication.) You also need to show that this really extends real multiplication, i.e. $av$ = $(a+0i)v$ where the left hand side term is to be understood as real multiplication as before. As an abelian group, $V$ is still the same, you just extended its “scalar field”. Then you can show that $T((a+ib)v) = (a+ib)T(v)$, meaning that $T$ is not only $ℝ$-linear, but also $ℂ$-linear.