Vector Subspace: Is W a Subspace of V?

[bugatti79]

Homework Statement

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

Homework Equations

I got this far

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

I believe I need to show that the constant 0 sequence has only finitely many non zero terms. My attempt

$W=\{x_1+y_1, x_2+y_2,...x_n+y_n,0,0 \}= Ʃ^{n}_{i=1} (x_n+y_n)$


Then I believe I need to show that $cx_n$ has only finitely many non zero terms if $x_n$ has...?

Any help will be appreciated. Thanks

PS. Where is the [itex] [/itex] tag?

 

[HallsofIvy]

Homework Statement

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

Homework Equations

I got this far

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

Yes, that is good.

I believe I need to show that the constant 0 sequence has only finitely many non zero terms. My attempt

$W=\{x_1+y_1, x_2+y_2,...x_n+y_n,0,0 \}= Ʃ^{n}_{i=1} (x_n+y_n)$

The "constant 0 sequence" has 0 non-zero terms- that's certainly finite!

Then I believe I need to show that $cx_n$ has only finitely many non zero terms if $x_n$ has...?

c(0)= 0 for any c so if $x_n$ has only 0 for n> N, $cx_n$ has only 0 for n> N.

Any help will be appreciated. Thanks

PS. Where is the [itex] [/itex] tag?

 

[bugatti79]

The "constant 0 sequence" has 0 non-zero terms- that's certainly finite!

Not sure I understand what you are saying here. Is my equation correct

c(0)= 0 for any c so if $x_n$ has only 0 for n> N, [itex]cx_n[/tex] has only 0 for n> N

So is this correct...

$c(x+y)=c(x_k+y_k) \in W$

 

[bugatti79]

IF we let c=0 then the above equation becomes

$0*(x+y)=0*(x_x+y_k) \in W$


Anyone willing to shed light on this simple problem for me?

Thanks

 

[kru_]

I am not sure why you want to use multiple elements to show closure of scalar multiplication.

If $x \in W$ then $x = x_n$ where $x_n$ has finitely many nonzero terms. Multiplying each term in $x_n$ by some scalar c doesn't change the number of nonzero terms. Right?


Next. Think about what the additive identity looks like. What sequence can be added to any other sequence without changing the number of nonzero terms?

 

[bugatti79]

I am not sure why you want to use multiple elements to show closure of scalar multiplication.

If $x \in W$ then $x = x_n$ where $x_n$ has finitely many nonzero terms. Multiplying each term in $x_n$ by some scalar c doesn't change the number of nonzero terms. Right?




Next. Think about what the additive identity looks like. What sequence can be added to any other sequence without changing the number of nonzero terms?

Could the 0 sequance be used?

$(0,0,0...)+(x_1+y_1, x_2+y_2, x_n+y_n,0,0)=(x_1+y_1, x_2+y_2, x_n+y_n,0,0)$

 

[HallsofIvy]

No, that's not enough. You have to show that the sum of any two series, each with a finite number of non-0 terms, has only a finite number of terms. Now, it is true that, if a series, $\{a_n}$ has a finite number of non-zero terms, there is some N such that if n> N, $a_n= 0$ ($a_n$ might be 0 for some n< N but that's not relevant). Another series, $\{b_n\}$, has all $b_n= 0$ for n> M, say. Do you see that both $a_n$ and $b_n$ are 0 for n> maximum(M, N)?

 

[bugatti79]

No, that's not enough. You have to show that the sum of any two series, each with a finite number of non-0 terms, has only a finite number of terms. Now, it is true that, if a series, $\{a_n}$ has a finite number of non-zero terms, there is some N such that if n> N, $a_n= 0$ ($a_n$ might be 0 for some n< N but that's not relevant). Another series, $\{b_n\}$, has all $b_n= 0$ for n> M, say. Do you see that both $a_n$ and $b_n$ are 0 for n> maximum(M, N)?

Hi HallsofIvy,

Yes, I believe I understand the sum part as this is what I showed in post 1.
You replied in post 2 regarding my 2 remaining questions on to show

1) the constant 0 sequence has only finitely many non zero terms

2) show that cxn has only finitely many non zero terms if xn

but I didnt quite understand what you were saying. Could you clarify a bit more?

Thanks

 

[bugatti79]

So my final attempt at putting it all together except for part 1)

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

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

1) show the constant 0 sequence has only finitely many non zero terms

2) show that $cx_n$ has only finitely many non zero terms if $x_n$

For any $c \in \mathbb{W}$ and if $x_n=0$ for n> N, then $cx_n=0$ for n> N


Thanks

 

[kru_]

You might want to briefly explain the reason why the 0 sequence has finitely many nonzero terms, but otherwise I think it looks ok.

 

[bugatti79]

You might want to briefly explain the reason why the 0 sequence has finitely many nonzero terms, but otherwise I think it looks ok.

I know the 0 sequence is (0,0,0,0...) but I don't know how to explain it has 'finitely many non 0 terms'.....

 

[HallsofIvy]

Well, how many non-zero numbers does it have? Isn't that a finite number?

 

[bugatti79]

it has 0 'non zero' numbers in it. but I didnt know that the number 0 is a finite number. Isnt that an undefined issue?

Other than that. I am happy with this thread for an answer :-)