# Linear Algebra vector spaces

1. Apr 18, 2009

### gotmilk04

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
I don't know what complex conjugation means, so I have no idea where to start with this.

2. Apr 18, 2009

### Tom Mattson

Staff Emeritus
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.

3. Apr 19, 2009

### 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)?

4. Apr 21, 2009

### Tom Mattson

Staff Emeritus
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).