Is A a Linear Subspace of ℝ^N?

  • Thread starter damabo
  • Start date
  • Tags
    Subspace
In summary: Therefore L is surjective.So L is a linear bijection, and its inverse is also linear, so L is an isomorphism. I think you are not being quite specific enough. You need to prove that ifL(a) = L(b)for some a, b in A, then a = b. You can't just argue that if the coefficients are different, the polynomials are different, because that's not necessarily true. You need to use the fact that the basis is linearly independent.
  • #1
damabo
54
0

Homework Statement



we have the vector space (ℝ,ℝ^N,+) of all sequences in ℝ. if A={(x_n) [itex]\in[/itex] ℝ^N | only finitely many components x_j differ from 0}, show that A is a linear subspace of ℝ^N. With which other vector space is this subspace isomorphous?

Homework Equations





The Attempt at a Solution


For the first question I would prove three criteria:1. A [itex]!= [/itex] ∅
2. A [itex]\subset[/itex] ℝ^N
3. if r,s [itex]\in[/itex] ℝ and (x_n),(y_n)[itex]\in[/itex] ℝ^N.
1. take for instance the sequence (x_n), where x=0 for all n [itex]\in[/itex] N. clearly (x_n)[itex]\in[/itex] A. So A !=∅.
2. choose (x_n) in A. then (x_n) [itex]\in[/itex] ℝ^N. So, A[itex]\subset[/itex] ℝ^N.
3. choose r,s in ℝ and (x_n),(y_n) in A. because only finitely many components x_j (and also y_j) will differ from 0, I would expect that from certain n_0, all x_n=0 for n ≥ n_0. and for certain n_1, all y_n=0 for n≥n_1. Thus if we choose any n≥n_2=max{n_0,n_1}, we will find that both x_n=y_n=0 so that rx_n + sy_n=0 for all n≥n_2. this means that r(x_n) + s(y_n)[itex]\in[/itex] A.

On the second question, I would try to find a subspace with the same dimension. But first I would have to find a basis for (ℝ,A,+) to reveal how many dimensions it has. here is where I am stuck.

thanks
 
Physics news on Phys.org
  • #2
Your proof that A is a subspace looks fine.

Regarding the dimension of A, you shouldn't need to find an explicit basis in order to answer this question: can dim(A) be finite?
 
  • #3
damabo said:

Homework Statement



we have the vector space (ℝ,ℝ^N,+) of all sequences in ℝ. if A={(x_n) [itex]\in[/itex] ℝ^N | only finitely many components x_j differ from 0}, show that A is a linear subspace of ℝ^N. With which other vector space is this subspace isomorphous?

That's "isomorphic". Hint: Does anything about ##A## remind you of polynomials?
 
  • #4
I notice that there are 'almost' [itex]\aleph[/itex] types of sequences.
0,0,0,...
x_1,0,0,0,...
x_1,x_2,0,0,0,...

and so on.

in any case, starting from some finite point n_0, there should be all zeroes. So its dimension would be n_0?
 
  • #5
damabo said:
I notice that there are 'almost' [itex]\aleph[/itex] types of sequences.
0,0,0,...
x_1,0,0,0,...
x_1,x_2,0,0,0,...

and so on.

in any case, starting from some finite point n_0, there should be all zeroes. So its dimension would be n_0?

Just to let you know, you don't actually need to know the dimension in order to solve this problem. LCKurtz has given you an excellent hint.

But it's certainly interesting and worthwhile to know what the dimension of A is. Here is a hint: A contains every sequence of the following form:

(1, 0, 0, 0, ...)
(0, 1, 0, 0, ...)
(0, 0, 1, 0, ...)
(0, 0, 0, 1, ...)
 
  • #6
so LCKurtz is saying that (ℝ,A,+) is isomorphic with (ℝ,ℝ[X],+), given that ℝ[X] are the polynomials of finite degree?
 
  • #7
damabo said:
so LCKurtz is saying that (ℝ,A,+) is isomorphic with (ℝ,ℝ[X],+), given that ℝ[X] are the polynomials of finite degree?

Yes, correct. Of course you need to state what exactly the mapping is between the two spaces, and prove that it's an isomorphism.
 
  • #8
ok thanks. will try to prove it tomorrow
 
  • #9
ok, I'll try to prove that (ℝ,A,+) is isomorphic with (ℝ,ℝ[X],+) if ℝ[x] stands for the polynomials of finite degree.
Define L: A -> ℝ[X] : a -> L(a).
1. L is bijective:
choose a in A. If α = {e_1 e_2, ..., e_n} is a standardbasis of A, then
a= a_1*e_1 + a_2*e_2 + ... +a_n*e_n is a unique linear combination with the basisvectors of α. If we then map those coefficients a_1,a_2,... onto the basis β={1,X,...,X^(n-1)}, so that L(a)=a_1+a_2*X+...+a_n*X^(n-1), we can see that L is a bijection, since the coefficients remain the same (L(a) is defined as a unique linear combination of the coefficients a_1,a_2, ... and its basis β).
2. L is linear:
choose λ,μ in ℝ en a,b in A. then L(λa+μb)=L((λa_1+μb_1)e_1+...+(λa_n+μb_n)) = (λa_1+μb_1)+(λa_2+μb_2)X+...+(λa_n+μb_n)X^(n-1)
which equals (λa_1+...+λa_nX^(n-1))+(μb_1+...+μb_nX^(n-1))=λ(a_1+...+a_nX^(n-1))+μ(b_1+...+b_nX^(n-1))
which of course equals λL(a)+μL(b)
 
  • #10
Your proof that L is linear looks fine. Your proof that L is bijective could use some clarification. Surjectivity is clear enough, but can you be more explicit in your argument that L is injective? Why are [itex]L(a_1 e_1 + \ldots + a_n e_n)[/itex] and [itex]L(b_1 e_1 + \ldots + b_n e_n)[/itex] different if [itex](a_1, \ldots, a_n) \neq (b_1, \ldots, b_n)[/itex]? In other words, how do we know that two polynomials with different coefficients can't still define the same function?
 
  • #11
L is injective:
because L is linear, L(a_1*e_1+...+a_n*e_n)=a_1*L(e_1) + ... + a_n L(e_n). similar goes for b_1,...,b_n.
Thus, if (a_1,...,a_n)!= (b_1,...,b_n), then L(a_1*e_1+...+a_n*e_n) != L(b_1*e_1+...+b_n*e_n), which means L is injective.

L is surjective:
choose a in A. if α={e_1,...,e_n} is a standardbasis for A, then a=x_1*e_1+...+x_n*e_n is a unique linear combination of the basisvectors. if we map a onto L(a)=x_1+...+x_n*X^(n-1), than L(a) is a unique linear combination of β={1, X, ...,X^(n-1)}.
is there, for every p in ℝ[X], an a in A such that L(a)=p?
choose p in ℝ[X]. p=a_1+a_2X+...+a_nX^(n-1), for coefficients a_1,...,a_n in ℝ. clearly, since span(A)=ℝ^n, a=a_1*e_1 + a_2*e_2 + ... +a_n*e_n [itex]\in[/itex] A. In this case, p=L(a), which is what we needed to prove.
 

What is a subspace?

A subspace is a subset of a vector space that also satisfies the properties of a vector space. In other words, it is a smaller space that has the same algebraic structure as the larger space.

What does it mean for two subspaces to be isomorphic?

Two subspaces are isomorphic if there exists a one-to-one and onto linear transformation between them. This means that the two subspaces have the same dimension and share all algebraic properties.

How do you determine if two subspaces are isomorphic?

To determine if two subspaces are isomorphic, you must check if they have the same dimension. If they do, then you must find a linear transformation that maps one subspace onto the other, while preserving all algebraic properties.

What is the importance of understanding isomorphism between subspaces?

Understanding isomorphism between subspaces is important in linear algebra, as it allows us to study different subspaces by studying their isomorphic counterparts. It also helps us to simplify and generalize complex problems by reducing them to simpler isomorphic subspaces.

Can two subspaces with different bases be isomorphic?

Yes, two subspaces can have different bases and still be isomorphic. This is because isomorphism is dependent on the underlying algebraic structure and not on the specific basis chosen.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
5K
  • Calculus and Beyond Homework Help
Replies
1
Views
493
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
Back
Top