Vector Subspaces

**bugatti79** · October 13th 2011, 10:12 PM

Hi Folks,

THis is my first exposure to Functional Analysis I am attempting to study and its proving a shock to my little brain. Here also is my first post which I hope is in the correct forum.

Let V be the vector space consisting of all infinite real sequences. Show that the subset W consisting of all such sequences with only finitely many non-0 entries is a subspace of V

If I write

$V^n={ (r_1, r_2, r_3) ; r_i \in R}$ for the vector V. Is this correct? Not sure how to represent the subset W.

$W^n={ (r_1, r_2, 0) ; r_{1,2} \in R}$ ?

Thanks

**FernandoRevilla** · October 13th 2011, 10:30 PM

Write (for example) $V=\{x=(x_n)_{n\geq 0}:x_n\in\mathbb{R}\}$ and $W=\{x\in V:\exists p\in\mathbb{N},\;x_k=0\textrm{\;if\;}k\geq p\}$

**Deveno** · October 13th 2011, 11:27 PM

fortunately you only need to show 3 things true for W:

1) the constant 0-sequence has only finitely many non-zero terms. (this is easy).
2) the sum of any two sequences with only finitely many non-zero terms has only finitely many non-zero terms.

(hint: any sequence $(x_n)$ with only finitely many non-zero terms has a maximim non-zero term, say $x_k$ . if $(y_n)$ is another such sequence, it has a maximum non-zero term, say $y_m$ . show that if t = max{k,m}, all $x_n+y_n$ for n > t are 0).

3) you must verify that the sequence $(cx_n)$ has only finitely many non-zero terms if $(x_n)$ does. this is also easy.

**bugatti79** · October 14th 2011, 07:45 AM

Originally Posted by Deveno

fortunately you only need to show 3 things true for W:

1) the constant 0-sequence has only finitely many non-zero terms. (this is easy).
2) the sum of any two sequences with only finitely many non-zero terms has only finitely many non-zero terms.

(hint: any sequence $(x_n)$ with only finitely many non-zero terms has a maximim non-zero term, say $x_k$ . if $(y_n)$ is another such sequence, it has a maximum non-zero term, say $y_m$ . show that if t = max{k,m}, all $x_n+y_n$ for n > t are 0).

3) you must verify that the sequence $(cx_n)$ has only finitely many non-zero terms if $(x_n)$ does. this is also easy.

1) I dont know what the constant 0 sequence is or why we need it?

2) If I let $x_n=(x_1, x_2, x_3....)$ and $y_n=(y_1, y_2, y_3...)$

Then $x_n+y_n=(x_1+y_1, x_2+y_2....)$

$x_n+y_n> 0$ if $n < t$ ,

$W= \{ x_n, y_n \in\mathbb {R}\}$

3) Let $c \in \mathbb{R} \therefore cx_n=c(x_1, x_2, x_3)$

This is the best I can do until I get some more guidance....thanks

**Deveno** · October 14th 2011, 08:45 AM

Originally Posted by bugatti79

1) I dont know what the constant 0 sequence is or why we need it?

2) If I let $x_n=(x_1, x_2, x_3....)$ and $y_n=(y_1, y_2, y_3...)$

Then $x_n+y_n=(x_1+y_1, x_2+y_2....)$

$x_n+y_n> 0$ if $n < t$ ,

$W= \{ x_n, y_n \in\mathbb {R}\}$

3) Let $c \in \mathbb{R} \therefore cx_n=c(x_1, x_2, x_3)$

This is the best I can do until I get some more guidance....thanks

you need to be sure W is not empty. the 0-vector is the traditional choice, because every vector space, no matter how small, must possess an additive identity.

you want to be sure the sum of two sequences who have only finitely many non-zero terms, is also such a sequence (this is called closure under vector addition). if only finitely many are non-zero, that means after a finite number, everything else is 0. here is one such sequence:

(x1,x2,x3,.......,xn,0,0,0,0,..........) (everything after xn is 0).

anothr such sequence might be:

(y1,y2,y3,......,ym,0,0,0,0,.........) (everything after ym is 0).

when we add them, we get (supposing n > m, which won't always happen)

(x1+y1,x2+y2,x3+y3,........,xm+ym,......,xn,0,0,0, 0,........) (everything after xn is 0)

your sequences are infinite, they never end (because the natural numbers never end).

this is why we represent them as (xj), where j = 1,2,3,4,5,....... going on forever.

the constant 0-sequence is the sequence whose terms are all 0:

(0,0,0,0,0,0,0,0,0,0,..................).

you can think of a sequence as a function f: j--->xj. so f(1) = x1, f(2) = x2, etc.

you can write these as a list (f(1),f(2),f(3),.........), but since the list goes on forever, you can't write it "all" down.

**FernandoRevilla** · October 14th 2011, 11:48 AM

An alternative for the sum (for example):

Let $x=(x_n)$ and $y=(y_n)$ be elements of $W$ , then there exits $p,q\in \mathbb{N}$ such that $x_k=0$ for all $k\geq p$ and $y_k=0$ for all $k\geq q$ . Choose $r=\max\{p,q\}$ , then $x_k+y_k=0$ for all $k\geq r$ , which implies $x+y=(x_k+y_k)\in W$ .

**bugatti79** · October 15th 2011, 12:16 AM

Originally Posted by FernandoRevilla

An alternative for the sum (for example):

Let $x=(x_n)$ and $y=(y_n)$ be elements of $W$ , then there exits $p,q\in \mathbb{N}$ such that $x_k=0$ for all $k\geq p$ and $y_k=0$ for all $k\geq q$ . Choose $r=\max\{p,q\}$ , then $x_k+y_k=0$ for all $k\geq r$ , which implies $x+y=(x_k+y_k)\in W$ .

Thank you Fernando and Deveno. I appreciate the help. Im slowly getting a feel for this topic.

1) Is the above answer acceptable for the question given ie, a mixture of words and math or must it all be math only?

2) How do you indicate that the subset W is a subspace of V?

Thanks

**Deveno** · October 15th 2011, 12:28 AM

the style of Fernando's response is fairly typical...symbols for precision (especially when describing elements, sets, and conditions), and words for smoother readability. if a subset W satisfies all 3 conditions laid out in post #3, it is a subspace. in words, you would say (having proven those conditions): "since W is a non-empty set closed under vector addition and scalar multiplication, it is a subspace of V". there is not, to my knowledge, any special symbol for "is a subspace of".

Fernando's post (#6) is a near-perfect demonstration of closure of vector addition, so that leaves you with 2 more conditions to verify.

**Hartlw** · October 15th 2011, 08:33 AM

Def: W is a subspace of V if for all x,y $\epsilon$ W and all a,b $\epsilon$ R, ax+by $\epsilon$ W.

0 does not belong to W so its not a subspace.

**Plato** · October 15th 2011, 08:44 AM

Originally Posted by Hartlw

Def: W is a subspace of V if for all x,y $\epsilon$ W and all a,b $\epsilon$ R, ax+by $\epsilon$ W.
0 does not belong to W so its not a subspace.

Are you saying that the sequence of all zeros does not contain only finitely many non-zero term?

**Hartlw** · October 15th 2011, 09:32 AM

From original post:
"Let V be the vector space consisting of all infinite real sequences. Show that the subset W consisting of all such sequences with only finitely many non-0 entries is a subspace of V"

A sequence of zeros does not have any non-zero entries, so it cannot have finiteley many of them. 0 does not belong to W.

**Plato** · October 15th 2011, 09:48 AM

Originally Posted by Hartlw

A sequence of zeros does not have any non-zero entries, so it cannot have finiteley many of them. 0 does not belong to W.

Now you are saying that zero is not a finite counting number.
Is that what you really mean to say.

Is this true, "A sequence that has no 1's in it, contains only finitely many 1's."?

**HallsofIvy** · October 15th 2011, 09:49 AM

Are you asserting that "0" is NOT a finite number? The 0 vector has NO non-zero entries and 0 certainly is finite.

**Hartlw** · October 15th 2011, 10:01 AM

Originally Posted by Plato

Now you are saying that zero is not a finite counting number.
Is that what you really mean to say.

Is this true, "A sequence that has no 1's in it, contains only finitely many 1's."?

This is what I am saying: A sequence of zeros does not have any non-zero entries, so it cannot have finiteley many of them. 0 does not belong to W.

If I have no books do I have a finite numbr of them?

What are you saying?

**Plato** · October 15th 2011, 10:03 AM

Originally Posted by Hartlw

If I have no books do I have a finite numbr of them?
What are you saying?

That is exactly what I am saying.
Zero is a finite number.

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