# Infinite dimensions

1. Apr 6, 2006

### stunner5000pt

Let V consists of all sequences [x0,x1x,...) of numbers and define vectors operations
$$[x_{0},x_{1},...) + [y_{0},y_{1},...) = (x_{0}+y_{0},...)$$
$$r[x_{0},x_{1},...) = [rx_{0},...}$$

SHow taht V is a vector space of infinite dimension

Well for some linear transformation T: V- >V

dim V = dim(ker T) + dim(im T)
ker T = {T(v) = 0, v in V}
i dont see how i can find the dimension of the ker or image for the matter...

Any ideas/suggestions?

2. Apr 6, 2006

### AKG

Why are you looking at linear transformations? And I'm pretty sure that formula you're using only works for finite dimensional V, otherwise it would give equations of the form $\infty = X + Y$. Either way, it's irrelevant to the question you've asked. First, just show that V is a vector space, i.e. that V under the given operations satisfies the vector space axioms. Then, show that V is not finite dimensional, by showing that no finite collection of vectors in V spans V. To do this, take an arbitrary finite collection of vectors, and construct a new vector that is not in the span of this collection.

3. Apr 7, 2006

### stunner5000pt

this arbitrary collection of vectors, should be in V? Since V is collection of all sequences of numbers, it shouldnt matter?

So form the question isnt
$$V = span[x_{0},x_{1},...)$$
our finite collection is
$$W = span\{v_{1},v_{2},...,v_{N}\} = r_{1} v_{1} + r_{2} v_{2} + ... r_{N} v_{N}$$
No matter what the N is, W cannot span V because V doesn't haev a finite N. WE are trying to span an infinite span with a finite span. So it isnt possible? Unless N - > infinity?

4. Apr 7, 2006

### HallsofIvy

Staff Emeritus
I think you are confusing the index denoting which sequence in a finite collection of sequences with the index denoting the term in a single sequence.

5. Apr 7, 2006

### stunner5000pt

so each of the v's are supposed to individual collections of numbers themselves??

6. Apr 7, 2006

### 0rthodontist

A somewhat more straightforward way to show that V is not finite dimensional is to show that for any natural number n, you can find n + 1 independent vectors in V, so V cannot be n-dimensional for any n.

7. Apr 7, 2006

### stunner5000pt

well ok then suppose
$$V = span\{v_{1},v_{2},...,v_{N}\}$$
also
$$r_{1} v_{1} + r_{2} v_{2} + ... r_{N} v_{N}= 0$$
and the r i are zero

so far so good?
since V is a collection of all possible numbers there must be exist an [itex] v_{n +1} [/tex] st vn+1 is in V
but how do i prove that v n+1 is not in the span??

8. Apr 7, 2006

### AKG

No. What do you mean the ri are zero? Do you just mean to say that the vi are linearly independent? V is not a collection of all possible numbers, V is a collection of sequences. And what do you mean that "there must exist a vn+1 such that vn+1 is in V"? That sentence there is essentially meaningless, it says nothing more than that V is nonempty. Here are three sequences:

v = [1, 2, 3, 5, 0, -1, 2, 1, 2, 1, 2, 1, 2, 1, ....)
w = [0, 0, 1, 2, 1, 1, 1, 2.3, 8.99, 8.99, 8.99, 8.99, 8.99, ...)
u = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ...)

Let a = 1, b = 1, c = 2. What is av + bw + cu? Just tell me the first 4 entries in the answer. Now, suppose I tell you that I have some constants a', b', and c' such that a'v + b'w + c'u has the same first three entries as av + bw + cu (which is just v + w + 2u). What can you say about the fourth entry in a'v + b'w + c'u?

9. Apr 7, 2006

### stunner5000pt

so
v + w + 2u = (5,8,14,21,...)

so for a'v + b' w + c'u the fourth entry may or may not be different?

10. Apr 8, 2006

### AKG

It may not be different? How so? How about this: I've told you that a'v + b'w + c'u also has first three entries 5, 8, and 14. Here, the v, w, and u are the same as before, but my a', b', and c' might be different. See if you can solve for a', b', and c'. See if they really can be different. If not, then the fourth entry must be the same.

11. Apr 8, 2006

### stunner5000pt

it can't be different because the a',b,',c' are all unique. I dunno what i was thinkin when i was saying taht th3ey may not be different

12. Apr 8, 2006

### AKG

What do you mean, "the a', b', c' are all unique"? The point is that if av + bw + cu and a'v + b'w + c'u have the same first three entries, then ALL their entries are the same, and a' = a, b' = b, and c' = c, assuming that {v,w,u} is a linearly independent basis. Can you prove this? If so, what can you say about the possibility of v, w, and u spanning V? Generalize this.