Is V a Vector Space over the Field of Real Numbers?

[gotmilk04]

Homework Statement

Let V be the set of all complex-valued functions, f, on the real line such that
f(-t)= f(t) with a bar over it, which denotes complex conjugation.
Show that V, with the operations
(f+g)(t)= f(t)+g(t)
(cf)(t)=cf(t)
is a vector space over the field of real numbers.

Homework Equations

The Attempt at a Solution

I don't know what complex conjugation means, so I have no idea where to start with this.

 

[quantumdude]

Let x,y=\in\mathbb{R}. Then z=x+iy\in\mathbb{C}. The complex conjugate of z is \overline{z}=x-iy.

So you have a function f:\mathbb{R}\rightarrow\mathbb{C} such that f(-t)=\overline{f(t)}. You can rewrite f(t) as f(t)=u(t)+iv(t), where u,v are real valued functions of a real variable t.

Chew on that and see if you don't know how to start.

 

[gotmilk04]

I'm still not sure what to do. I have to show all the properties of a vector space, right?
But I'm not sure how to write it out and everything.
Like, for vector addition, do I add another fuction g(t)=x(t)+iy(t) where x,y are real valued and then add f(t)+g(t) and show it equals g(t)+f(t)?

 

[quantumdude]

Yes, you have to show that all the properties of a vector space hold. That f(t)+g(t)=g(t)+f(t) is obvious because the functions are complex valued. The real heart of the matter (as far as vector addition goes) is showing that f(t)+g(t) is even IN the vector space. That is, you have to show that (f+g)(-t)=(f+g)(t)-bar (sorry, Latex isn't working).