# Vector Subspace Question

1. Oct 16, 2011

### bugatti79

1. The problem statement, all variables and given/known data

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

2. Relevant 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?

2. Oct 16, 2011

### HallsofIvy

Yes, that is good.

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

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

Last edited by a moderator: Oct 18, 2011
3. Oct 17, 2011

### bugatti79

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

So is this correct...

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

4. Oct 17, 2011

### 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

5. Oct 17, 2011

### 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?

6. Oct 18, 2011

### bugatti79

Could the 0 sequance be used?

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

7. Oct 18, 2011

### 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)?

8. Oct 18, 2011

### bugatti79

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

9. Oct 19, 2011

### 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

10. Oct 19, 2011

### 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.

11. Oct 20, 2011

### bugatti79

I know the 0 sequence is (0,0,0,0....) but I dont know how to explain it has 'finitely many non 0 terms'..............

12. Oct 20, 2011

### HallsofIvy

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

13. Oct 20, 2011

### 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. Im happy with this thread for an answer :-)

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