# Which subspace is isomorf

damabo

## Homework Statement

we have the vector space (ℝ,ℝ^N,+) of all sequences in ℝ. if A={(x_n) $\in$ 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?

## The Attempt at a Solution

For the first question I would prove three criteria:1. A $!=$ ∅
2. A $\subset$ ℝ^N
3. if r,s $\in$ ℝ and (x_n),(y_n)$\in$ ℝ^N.
1. take for instance the sequence (x_n), where x=0 for all n $\in$ N. clearly (x_n)$\in$ A. So A !=∅.
2. choose (x_n) in A. then (x_n) $\in$ ℝ^N. So, A$\subset$ 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)$\in$ 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

Homework Helper
Gold Member
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?

Homework Helper
Gold Member

## Homework Statement

we have the vector space (ℝ,ℝ^N,+) of all sequences in ℝ. if A={(x_n) $\in$ 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?

damabo
I notice that there are 'almost' $\aleph$ 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?

Homework Helper
Gold Member
I notice that there are 'almost' $\aleph$ 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, ...)

damabo
so LCKurtz is saying that (ℝ,A,+) is isomorphic with (ℝ,ℝ[X],+), given that ℝ[X] are the polynomials of finite degree?

Homework Helper
Gold Member
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.

damabo
ok thanks. will try to prove it tomorrow

damabo
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)

Homework Helper
Gold Member
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 $L(a_1 e_1 + \ldots + a_n e_n)$ and $L(b_1 e_1 + \ldots + b_n e_n)$ different if $(a_1, \ldots, a_n) \neq (b_1, \ldots, b_n)$? In other words, how do we know that two polynomials with different coefficients can't still define the same function?

damabo
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 $\in$ A. In this case, p=L(a), which is what we needed to prove.