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 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 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, cx_n[/tex] has only 0 for n&gt; N <br />

<br /> <br /> So is this correct...<br /> <br /> 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 :-)