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)

[itex] f(-t) = \overline{f(t)}[/itex] 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)

[itex] f(-t) = \overline{f(t)}[/itex] 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 [itex]f(x)= 3x- x^3i[/itex]. More generally, f(x)= g(x)+ h(x)i where g and h are real valued functions of x.

The requirement that [itex]f(-t)= \overline{f(t)}[/itex] 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 [itex]f(x)= 3x- x^3i[/itex]. 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 [itex]f(-t)= \overline{f(t)}[/itex] 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)?