# Complex Conjugate Vector Space

1. Sep 6, 2014

### dkotschessaa

I'm confused about some of the notation in Hoffman & Kunze Linear Algebra.

Let V be the set of all complex valued functions f on the real line such that (for all t in R)

$f(-t) = \overline{f(t)}$ where the bar 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. Give an example of a function in V which is not real valued.

Before I can even approach the question I need to clarify what is happening here.

I'm not sure why it is written as $f(-t) = \overline{f(t)}$

Am I correct that $f(a) = a - 0i = a$ (since a is real).
$f(-a)$ would be $a + 0i = a$

Obviously I have some confusion here.

Also, with the addition properties given above, would I not have:

$(f+g)(t) = f(t) + g(t) = \overline{f(-t)} + \overline{g(-t)}$

In particular I am confused when I try to show closure under the real numbers. since $(a_{1} + b_{1}i) + (a_{2} + b_{2}i)$ gives me (I think) something like $a_{1} + a_{2} + (b_{1} + b_{2})i$

Unless I'm supposed to be adding something to it's OWN conjugation, in which case the imaginary part would cancel, which would be nice. Is that what I am supposed to do?

Appreciate any help

-Dave K

Last edited: Sep 6, 2014
2. Sep 6, 2014

### HallsofIvy

Staff Emeritus
No, you are not correct. If f(x) is NOT necessarily equal to x! For any real x, f(x) is some complex number depending upon x. It might be, for example, f(x)= xi or $f(x)= 3x- x^3i$. More generally, f(x)= g(x)+ h(x)i where g and h are real valued functions of x.

The requirement that $f(-t)= \overline{f(t)}$ means that g(-t)= g(t) and h(-t)= -h(t). That is, g is an "even" function and h is an "odd" function.

3. Sep 6, 2014

### dkotschessaa

I meant that if a is just some integer, for example. What is f(a)? a = a + 0i, whose conjugate is a = 0i,which is a. Perhaps my thinking is not function-oriented enough here.

That sheds a bit more light, but I'm still confused about closure.

If I have $f(-x)= g(x)+ h(x)i$ and $f'(x) = g'(x)+ h'(x)i$ I still wind up with something with an imaginary part when I add them.

-Dave K

4. Sep 6, 2014

### dkotschessaa

Am I only taking the conjugate when I have $f(-t)$ and not when I have just $f(t)$ as is originally defined? i.e. $f(-t) = \overline{f(t)}$ but $f(t) = f(t)$ (not conjugated)?