Complex Conjugate Vector Space

[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

 

[HallsofIvy]

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 $a + 0i = a$

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.

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

 

[dkotschessaa]

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.

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.

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.

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

 

[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)?