# Linear Algebra vector spaces

## 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.

## The Attempt at a Solution

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

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Tom Mattson
Staff Emeritus
Gold Member
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.

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)?

Tom Mattson
Staff Emeritus