Complex periodic functions in a vector space

  • #26
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,136
8,947
Qc) Suppose f(t) belongs to V. For a complex number c, show that h(t) =cf(t) also belongs to V.

Answer

Let f be a periodic function of period 1 and h =cf where c is a complex number.

∀t h(t+1) =cf(t+1) =cf(t) = h(t)

Hence h is periodic with period 1 QED.

Yes, that's it.

Qd)
Is V with the above addition and scaling operations a vector space?

Would I need to go through the axioms to prove that V was definitely a vector space?

It depends. If you can assume that the set of all complex functions is a vector space, then you need only show that addition and scalar multiplication are closed on the subset you are dealing with (in this case functions with period 1).

This is true for vector spaces in general:

If ##U## is a subset of a vector space ##V## and:

##u, v \in U## and ##c \in \mathbb{C} \ \Rightarrow u + v \in U## and ##cu \in U##

then ##U## is a vector subspace. Most axioms are met directly because the axioms hold for ##V##, but you might like to think about why the above ensures that the ##0## vector is in ##U## and also why it ensures that ##u \in U \Rightarrow -u \in U##. And, you might like to look at the other axioms as well and convince yourself why they must be met for any subset of a vector space.

With that assumption, you have shown that the set of functions of period 1 is a vector space.
 
  • #27
122
4
Yes, that's it.



It depends. If you can assume that the set of all complex functions is a vector space, then you need only show that addition and scalar multiplication are closed on the subset you are dealing with (in this case functions with period 1).

This is true for vector spaces in general:

If ##U## is a subset of a vector space ##V## and:

##u, v \in U## and ##c \in \mathbb{C} \ \Rightarrow u + v \in U## and ##cu \in U##

then ##U## is a vector subspace. Most axioms are met directly because the axioms hold for ##V##, but you might like to think about why the above ensures that the ##0## vector is in ##U## and also why it ensures that ##u \in U \Rightarrow -u \in U##. And, you might like to look at the other axioms as well and convince yourself why they must be met for any subset of a vector space.

With that assumption, you have shown that the set of functions of period 1 is a vector space.

Is it the case that as the addition and scaling axioms are present in U , these can then be used to find to -u by scaling by a factor of (-1) and the 0 vector can found by summing u + (-u) = 0 ?
 
  • #28
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,136
8,947
Is it the case that as the addition and scaling axioms are present in U , these can then be used to find to -u by scaling by a factor of (-1) and the 0 vector can found by summing u + (-u) = 0 ?

Not quite. The zero vector is in ##U## because ##0u = 0 \in U## for ##u \in U##.

If you were a pure maths student, at this point you might jump up and point out that means you need at least one ##u \in U##. And, therefore, ##U## must be a non-empty subset of ##V## in order to be a vector space.
 
  • #29
122
4
Not quite. The zero vector is in ##U## because ##0u = 0 \in U## for ##u \in U##.

If you were a pure maths student, at this point you might jump up and point out that means you need at least one ##u \in U##. And, therefore, ##U## must be a non-empty subset of ##V## in order to be a vector space.

Is it a fair assumption to make to say that the set of all complex functions is a vector space?
 

Related Threads on Complex periodic functions in a vector space

Replies
2
Views
5K
Replies
5
Views
15K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
705
Replies
3
Views
2K
Replies
17
Views
790
Replies
3
Views
7K
  • Last Post
Replies
4
Views
2K
Top