[Linear Algebra] Another question on subspaces

[iJake]

Homework Statement

Let $V$ be the vector space of the sequences which take real values. Prove whether or not the following subsets $W \in V$ are subspaces of $(V, +, \cdot)$

a) $W = \{(a_n) \in V : \sum_{n=1}^\infty |a_n| < \infty\}$

b) $W = \{(a_n) \in V : \lim_{n\to \infty} a_n = 1\}$

c) $W = \{(a_n) \in V : ∃ \lim_{n\to \infty} a_2n\}$

d) $W = \{(a_n) \in V : (cardinality) \{n : a_n \neq 0\} < \infty \}$

The # for my cardinality in the last subspace was breaking my formatting.

Homework Equations

The three conditions to be a subspace under $V$ are that vector addition and scalar multiplication be closed, and that the 0 vector belongs to the subspace.

The Attempt at a Solution

[/B]
The notation and logic here is throwing me off a bit because I have not yet begun to study Calculus and I am unfamiliar with the operations and behavior of sequences.

For a)

$a_{n+2} + b_{n+2} = (a_{n+1} + b_{n+1}) + (a_n + b_n)$
$\lambda \cdot (a_n, a_{n+1}, a_{n+2}) = (\lambda a_n, \lambda a_{n+1}) \lambda a_{n+2}) \forall \lambda \in \mathbb k , \forall a_n \in W$
if $a_n = 0$, the sequence is 0, and $b_n + 0 = b_n$
$0 \cdot a_n = 0 \forall a_n \in W$

I conclude that in a) W is a subspace.

In b), I recognize that there is a problem with the 0 vector, but I'm not sure how to describe it in notation. I don't think that b) is a subspace.

For c) I think that all of the properties should hold, but I am again confused about how to write out the notation. For example, to prove closure under addition, should I be adding limits? For closure under multiplication, multiplying scalars by limits? Sorry if this is a bit of a silly question, but I just don't know how to write it out. I feel like all of the properties should hold, however.

In d) there is again a problem with the 0 vector, in that $a_n = 0 ∉ W$, thus I conclude d) is not a subspace.

All help is greatly appreciated.

 

[fresh_42]

Homework Statement

Let $V$ be the vector space of the sequences which take real values. Prove whether or not the following subsets $W \in V$ are subspaces of $(V, +, \cdot)$

a) $W = \{(a_n) \in V : \sum_{n=1}^\infty |a_n| < \infty\}$

b) $W = \{(a_n) \in V : \lim_{n\to \infty} a_n = 1\}$

c) $W = \{(a_n) \in V : ∃ \lim_{n\to \infty} a_2n\}$

d) $W = \{(a_n) \in V : (cardinality) \{n : a_n \neq 0\} < \infty \}$

The # for my cardinality in the last subspace was breaking my formatting.

Homework Equations

The three conditions to be a subspace under $V$ are that vector addition and scalar multiplication be closed, and that the 0 vector belongs to the subspace.

The Attempt at a Solution

[/B]
The notation and logic here is throwing me off a bit because I have not yet begun to study Calculus and I am unfamiliar with the operations and behavior of sequences.

For a)

$a_{n+2} + b_{n+2} = (a_{n+1} + b_{n+1}) + (a_n + b_n)$
$\lambda \cdot (a_n, a_{n+1}, a_{n+2}) = (\lambda a_n, \lambda a_{n+1}) \lambda a_{n+2}) \forall \lambda \in \mathbb k , \forall a_n \in W$
if $a_n = 0$, the sequence is 0, and $b_n + 0 = b_n$
$0 \cdot a_n = 0 \forall a_n \in W$

I conclude that in a) W is a subspace.

Yes.

In b), I recognize that there is a problem with the 0 vector, but I'm not sure how to describe it in notation. I don't think that b) is a subspace.

Yes, it is no subspace for the reason you mentioned. You could write is as $0_V=(0,0,0,\ldots) \notin W$ because $\lim_{n\to \infty}0_V=0$. But just saying $0_V\notin W$ will do.

For c) I think that all of the properties should hold, but I am again confused about how to write out the notation. For example, to prove closure under addition, should I be adding limits?

Probably the shortest way to show the existence of limits $\lim_{n \to \infty}(a_{2n}+b_{2n}).$
$$\exists \,A,B\, : \,\lim_{n \to \infty}a_{2n}=A\, , \,\lim_{n \to \infty}b_{2n}=B \Longrightarrow \lim_{n \to \infty}(a_{2n}+b_{2n})=A+B \Longrightarrow \exists \,\lim_{n \to \infty}(a_{2n}+b_{2n})$$

For closure under multiplication, multiplying scalars by limits? Sorry if this is a bit of a silly question, but I just don't know how to write it out. I feel like all of the properties should hold, however.

... and similar for the scalar multiples.
$$
\forall \,\lambda \in \mathbb{R}\, : \,\lim_{n \to \infty}a_{2n}=A \Longrightarrow \lim_{n \to \infty}\lambda(a_{2n})=\lambda \lim_{n \to \infty}a_{2n}=\lambda A\in \mathbb{R} \Longrightarrow \exists\, \lim_{n \to \infty}\lambda (a_{2n})
$$

In d) there is again a problem with the 0 vector, in that $a_n = 0 ∉ W$, thus I conclude d) is not a subspace.

What is $| \{ \, n \, : \, 0_n \neq 0 \, \} | = (cardinality) \{\,n\,:\,0_n\neq 0\,\}\,?$

 

[member 587159]

Yes.Yes, it is no subspace for the reason you mentioned. You could write is as $0_V=(0,0,0,\ldots) \notin W$ because $\lim_{n\to \infty}0_V=0$. But just saying $0_V\notin W$ will do.
Probably the shortest way to show the existence of limits $\lim_{n \to \infty}(a_{2n}+b_{2n}).$
$$\exists \,A,B\, : \,\lim_{n \to \infty}a_{2n}=A\, , \,\lim_{n \to \infty}b_{2n}=B \Longrightarrow \lim_{n \to \infty}(a_{2n}+b_{2n})=A+B \Longrightarrow \exists \,\lim_{n \to \infty}(a_{2n}+b_{2n})$$
... and similar for the scalar multiples.
$$
\forall \,\lambda \in \mathbb{R}\, : \,\lim_{n \to \infty}a_{2n}=A \Longrightarrow \lim_{n \to \infty}\lambda(a_{2n})=\lambda \lim_{n \to \infty}a_{2n}=\lambda A\in \mathbb{R} \Longrightarrow \exists\, \lim_{n \to \infty}\lambda (a_{2n})
$$

What is $| \{ \, n \, : \, 0_n \neq 0 \, \} | = (cardinality) \{\,n\,:\,0_n\neq 0\,\}\,?$

For (a) don't you think there is something wrong with his/her reasoning? Or maybe I don't understand the notation.

For (d) he/she means the set of sequences with finite support.

 

[Mark44]

Homework Statement

Let $V$ be the vector space of the sequences which take real values. Prove whether or not the following subsets $W \in V$ are subspaces of $(V, +, \cdot)$

a) $W = \{(a_n) \in V : \sum_{n=1}^\infty |a_n| < \infty\}$

b) $W = \{(a_n) \in V : \lim_{n\to \infty} a_n = 1\}$

c) $W = \{(a_n) \in V : ∃ \lim_{n\to \infty} a_2n\}$

Do you mean $W = \{(a_n) \in V : ∃ \lim_{n\to \infty} a_{2n}\}$?
In LaTeX, if subscripts, exponents, fractions, etc. are more than 1 character, you have to surround them with braces.
If so, another way to write the last limit is $\lim_{n\to \infty} a_{2n} < \infty\}$
Could we have a sequence where the limit exists for the even-subscript terms, but not for the odd-subscript terms?

d) $W = \{(a_n) \in V : (cardinality) \{n : a_n \neq 0\} < \infty \}$

To display the # character, add a backslash just in front of it, like this ##\# ##
Think about what your set W means in terms of the number of nonzero terms in the sequence vs. the number of zero terms. Is there a 0 sequence in W? If you have two such sequences, is their sum in W? Are scalar multiples of such a sequence also in W.

The # for my cardinality in the last subspace was breaking my formatting.

Homework Equations

The three conditions to be a subspace under $V$ are that vector addition and scalar multiplication be closed, and that the 0 vector belongs to the subspace.

The Attempt at a Solution

[/B]
The notation and logic here is throwing me off a bit because I have not yet begun to study Calculus and I am unfamiliar with the operations and behavior of sequences.

For a)

$a_{n+2} + b_{n+2} = (a_{n+1} + b_{n+1}) + (a_n + b_n)$
$\lambda \cdot (a_n, a_{n+1}, a_{n+2}) = (\lambda a_n, \lambda a_{n+1}) \lambda a_{n+2}) \forall \lambda \in \mathbb k , \forall a_n \in W$
if $a_n = 0$, the sequence is 0, and $b_n + 0 = b_n$
$0 \cdot a_n = 0 \forall a_n \in W$

I conclude that in a) W is a subspace.

I agree with your conclusion, but I'm confused by what you're saying in the work supporting your conclusion.

In b), I recognize that there is a problem with the 0 vector, but I'm not sure how to describe it in notation. I don't think that b) is a subspace.

For c) I think that all of the properties should hold, but I am again confused about how to write out the notation. For example, to prove closure under addition, should I be adding limits? For closure under multiplication, multiplying scalars by limits? Sorry if this is a bit of a silly question, but I just don't know how to write it out. I feel like all of the properties should hold, however.

In d) there is again a problem with the 0 vector, in that $a_n = 0 ∉ W$, thus I conclude d) is not a subspace.

All help is greatly appreciated.

 

[fresh_42]

For (a) don't you think there is something wrong with his/her reasoning? Or maybe I don't understand the notation.

You're right. I confused the cases and only looked at the result, and disregarded the details, esp. the lack of the triangle in equality.

For (d) he/she means the set of sequences with finite support.

I know. That was a rhetoric question.

 

[iJake]

es, it is no subspace for the reason you mentioned. You could write is as $0_V=(0,0,0,\ldots) \notin W$ because $\lim_{n\to \infty}0_V=0$. But just saying $0_V\notin W$ will do.

That should be $\lim_{n\to \infty}0_V=1$, correct?

What is $| \{ \, n \, : \, 0_n \neq 0 \, \} | = (cardinality) \{\,n\,:\,0_n\neq 0\,\}\,?$

Perhaps my interpretation is incorrect, but I believe this refers to a finite set of sequences.

My reasoning is that if the cardinality of $a_n$ is $\#\{n : a_n \neq 0\}$ then there is no $a_n = 0 \in W$.

For (a) don't you think there is something wrong with his/her reasoning? Or maybe I don't understand the notation.

Hello, could you elaborate please? :) Thank you for replying.

For (d) he/she means the set of sequences with finite support.

I am not precisely sure what 'support' means in this context, but I believe your statement agrees with mine just above this?

Do you mean $W = \{(a_n) \in V : ∃ \lim_{n\to \infty} a_{2n}\}$?
In LaTeX, if subscripts, exponents, fractions, etc. are more than 1 character, you have to surround them with braces.
If so, another way to write the last limit is $\lim_{n\to \infty} a_{2n} < \infty\}$
Could we have a sequence where the limit exists for the even-subscript terms, but not for the odd-subscript terms?

Yes, that is what I meant! Why does the existence of that limit guarantee that $\lim_{n\to \infty} a_{2n} < \infty$ holds? Sorry for not following that line of logic!

I only see that the sequence is defined for limits to exist for the terms with even subscripts, so I would say that the answer to your question is yes. But is that relevant? My line of reasoning, even though I was unable to translate it into notation, was along the lines of fresh_42's proof which seems to indicate that c) is a subspace.

To display the # character, add a backslash just in front of it, like this ##\# ##
Think about what your set W means in terms of the number of nonzero terms in the sequence vs. the number of zero terms. Is there a 0 sequence in W? If you have two such sequences, is their sum in W? Are scalar multiples of such a sequence also in W.

I agree with your conclusion, but not with the work supporting your conclusion.

I don't see a 0 sequence in W, but maybe I am not understanding the problem statement correctly.

Could you elaborate a little on the flaw in my proof for a)?

Thank you everyone for your help.

 

[fresh_42]

That should be $\lim_{n\to \infty}0_V=1$, correct?

No. How should we ever reach $1$ if we are stuck with $0's$? This is the reason why $W_{b)}$ isn't a subspace. It cannot be $1$ as required.

Perhaps my interpretation is incorrect, but I believe this refers to a finite set of sequences.
My reasoning is that if the cardinality of $a_n$ is $\{n : a_n \neq 0\}$ then there is no $a_n = 0 \in W$.

I think you have difficulties in understanding what the elements are. The vectors of $V$ are all sequences:
$$
a := (a_n)_{n\in \mathbb{N}} = (a_1,a_2,\ldots )
$$
This is one single vector. Two of them are added componentwise $a+b=(a_n)+(b_n)=(a_n+b_n)=(a+b)$, and the scalar multiples, too, $\lambda (a_n)_{n \in \mathbb{N}}=(\lambda a_n)_{n \in \mathbb{N}}=\lambda a\,.$ This makes the zero vector a sequence $0=(0,0,0,\ldots)$. Now count all the sequence elements which are different from zero and decide, whether your result is finite or not.

Support generally means: all locations different from zero. The exact definition might depend on the case.

You should now first try to correct your reasoning in case a).

 

[Mark44]

Do you mean $W = \{(a_n) \in V : ∃ \lim_{n\to \infty} a_{2n}\}$?
In LaTeX, if subscripts, exponents, fractions, etc. are more than 1 character, you have to surround them with braces.
If so, another way to write the last limit is $\lim_{n\to \infty} a_{2n} < \infty\}$
Could we have a sequence where the limit exists for the even-subscript terms, but not for the odd-subscript terms?

Yes, that is what I meant! Why does the existence of that limit guarantee that $\lim_{n\to \infty} a_{2n} < \infty$ holds? Sorry for not following that line of logic!

You can disregard my comment. I was mistakenly thinking that V consisted of convergent sequences. For part c, it doesn't matter what the odd-index sequence elements are doing.

 

[iJake]

$a_{n+2} + b_{n+2} = (a_{n+1} + b_{n+1}) + (a_n + b_n)$
$\lambda \cdot (a_n, a_{n+1}, a_{n+2}) = (\lambda a_n, \lambda a_{n+1}) \lambda a_{n+2}) \forall \lambda \in \mathbb k , \forall a_n \in W$
if $a_n = 0$, the sequence is 0, and $b_n + 0 = b_n$
$0 \cdot a_n = 0 \forall a_n \in W$

$a_{n+2} + b_{n+2} = (a_{n+1} + b_{n+1}) + (a_n + b_n)$
$\lambda \cdot (a_n) = (\lambda a_n)$

I think my proof for scalar multiplication at the very least looked a bit goofy.
I'm now looking at my 0 vector, but I think it holds? If $a_n = 0$ it is the additive identity, and $0 \cdot a_n$ returns the zero vector you described: $0=(0,0,0,\ldots)$

No. How should we ever reach $1$ if we are stuck with $0's$? This is the reason why $W_{b)}$ isn't a subspace. It cannot be $1$ as required.

Sorry, I was just misunderstanding the semantics. Of course we never reach 1, and for the condition to hold we would need to, thus the 0 vector doesn't belong. I understand this proof now.

I think you have difficulties in understanding what the elements are. The vectors of $V$ are all sequences:
$$
a := (a_n)_{n\in \mathbb{N}} = (a_1,a_2,\ldots )
$$
This is one single vector. Two of them are added componentwise $a+b=(a_n)+(b_n)=(a_n+b_n)=(a+b)$, and the scalar multiples, too, $\lambda (a_n)_{n \in \mathbb{N}}=(\lambda a_n)_{n \in \mathbb{N}}=\lambda a\,.$ This makes the zero vector a sequence $0=(0,0,0,\ldots)$. Now count all the sequence elements which are different from zero and decide, whether your result is finite or not.

Support generally means: all locations different from zero. The exact definition might depend on the case.

You should now first try to correct your reasoning in case a).

Do you mean to count all the sequence elements of the 0 vector which are different from 0? There are none. Is the meat of what we're looking at supposed to be the idea that there are infinite such sequence elements, and thus the condition for the subspace doesn't hold?

 

[fresh_42]

$a_{n+2} + b_{n+2} = (a_{n+1} + b_{n+1}) + (a_n + b_n)$
$lambda \cdot (a_n) = (\lambda a_n)$

I think my proof for scalar multiplication at the very least looked a bit goofy.

It all looks goofy. Where did you read this law of addition, where the sequence is recursively defined?
I assume we are talking about part a). Here we have two elements $a=(a_n)_{n\in \mathbb{N}}\; , \;b=(b_n)_{n\in \mathbb{N}}$ with $\sum_n |a_n | < \infty\; , \;\sum_n|b_n|< \infty$ and must show, that for $a+b$ and $\lambda a$ we have $\sum_n |a_n+b_n|< \infty$, resp. $\sum_n|\lambda a_n|<\infty$.

Do you mean to count all the sequence elements of the 0 vector which are different from 0? There are none.

Yes and yes.

Is the meat of what we're looking at supposed to be the idea that there are infinite such sequence elements, and thus the condition for the subspace doesn't hold?

I don't understand this sentence. If $\# \{\,n\, : \,a_n=0_n \neq 0\,\} = 0 < \infty$, how do you get at $0 \notin W_{d)}\,?$ $W_{d)}$ contains all sequences which have only finitely many elements different from zero. Now the question is: Will this change if we add two of them or multiply them with a scalar $\lambda \,?$ And zero is a finite number.

 

[iJake]

It all looks goofy. Where did you read this law of addition, where the sequence is recursively defined?
I assume we are talking about part a). Here we have two elements $a=(a_n)_{n\in \mathbb{N}}\; , \;b=(b_n)_{n\in \mathbb{N}}$ with $\sum_n |a_n | < \infty\; , \;\sum_n|b_n|< \infty$ and must show, that for $a+b$ and $\lambda a$ we have $\sum_n |a_n+b_n|< \infty$, resp. $\sum_n|\lambda a_n|<\infty$.

I must say I got a good chuckle out of "it all looks goofy." It looks like you've basically resolved this for me.

$\sum_n |a_n| + \sum_n |b_n| = \sum_n |a_n+b_n|< \infty$ $\forall a_n, b_n$
$\lambda \cdot \sum_n |a_n| = \sum_n|\lambda a_n| <\infty$ $\forall \lambda, \forall a_n$

I don't understand this sentence. If $\# \{\,n\, : \,a_n=0_n \neq 0\,\} = 0 < \infty$, how do you get at $0 \notin W_{d)}\,?$ $W_{d)}$ contains all sequences which have only finitely many elements different from zero. Now the question is: Will this change if we add two of them or multiply them with a scalar $\lambda \,?$ And zero is a finite number.

You're right, I wasn't fully grasping the idea till this post. I understand now that there are no sequence elements different from 0, so it holds that # is 0, which is finite. I now have my zero vector. Would the notation for the rest look something like:

$\# \{n : a_n \neq 0 \} + \# \{n : b_n \neq 0\} = \# \{n : (a_n + b_n) \neq 0\} \in W$

$\lambda \cdot \# \{n : a_n \neq 0 \} = \# \{n : \lambda a_n \neq 0 \} \in W$

Thanks for your patience when faced with my goofiness, fresh_42!

 

[fresh_42]

I must say I got a good chuckle out of "it all looks goofy." It looks like you've basically resolved this for me.

More important is whether you've understood why this must be shown.

$\sum_n |a_n| + \sum_n |b_n| = \sum_n |a_n+b_n|< \infty$ $\forall a_n, b_n$
$\lambda \cdot \sum_n |a_n| = \sum_n|\lambda a_n| <\infty$ $\forall \lambda, \forall a_n$



You're right, I wasn't fully grasping the idea till this post. I understand now that there are no sequence elements different from 0, so it holds that # is 0, which is finite. I now have my zero vector. Would the notation for the rest look something like:

$\# \{n : a_n \neq 0 \} + \# \{n : b_n \neq 0\} = \# \{n : (a_n + b_n) \neq 0\} \in W$

This equality doesn't hold. What if $a=(1,2,3,0,0,\ldots)$ and $b=(-1,-2,-3,0,0,\ldots)$ or $b=(0,0,0,1,-2,3,0,0,\ldots)\,?$ The specific cardinality of any set $\{n : a_n \neq 0 \}$ is completely irrelevant, so you don't have to compute what cannot be computed. Only that it is finite counts.

$\lambda \cdot \# \{n : a_n \neq 0 \} = \# \{n : \lambda a_n \neq 0 \} \in W$

And in case $\lambda=0\,?$

 

[iJake]

More important is whether you've understood why this must be shown.

Yes, of course, and I do.

This equality doesn't hold. What if $a=(1,2,3,0,0,\ldots)$ and $b=(-1,-2,-3,0,0,\ldots)$ or $b=(0,0,0,1,-2,3,0,0,\ldots)\,?$ The specific cardinality of any set $\{n : a_n \neq 0 \}$ is completely irrelevant, so you don't have to compute what cannot be computed. Only that it is finite counts.

Of course, I responded hastily and did not consider that, naturally, the equality doesn't hold if $b_n$ = $-1 \cdot a_n$. Nonetheless, the sum of two finite sets is a finite set, no? So it would look more like

$\# \{n : a_n \neq 0 \} < \infty + \# \{n : b_n \neq 0 \} < \infty = \# \{n : a_n \neq 0 \} + \# \{n : b_n \neq 0 \} < \infty$

And in case $\lambda=0\,?$

All of the elements of the sequence would be 0 and I would have the 0 vector, which is finite, and the same sum would be true.

 

[fresh_42]

Yes, only that it is finite is important. In your last equation I would replace the first $+$ sign by a comma, as you don't add the cardinalities. If at all, then $\infty > \# \{n\, : \,a_n \neq 0\,\} + \# \{n\, : \,b_n \neq 0\,\} \ge \# \{n\, : \,a_n+b_n \neq 0\,\}$

 

[iJake]

Thank you fresh_42, your help working through my exercises really helps to refine my understanding. I've been able to do many more exercises on my own than before.

 

[MidgetDwarf]

Also. What would make this exercise conceptually easier, would be understanding the properties of limits. Ie. The sum of two convergent sequences is convergent, multiplying a convergent sequence by a scalar...

Since for part b, any element of W converges to 1 (approaches the limiting value of 1), so any convergent sequence multiplied by a scaler is not an element of W. Therefore W is not closed under scalar multiplication.

If I were doing, this exercise (part B), i would just state the above rules. There is no need to actually write out the elements of an arbitrary element of W.