# Homework Help: Spanning Sets

1. Oct 24, 2004

### Spectre5

I don't totally understand spanning sets...

Can anyone explain this problem to me:

let V = set of all polynomials with degree of 2 or less (a vector space_
let S = {t + 1, t^2 + 1, t^2 - t}

Does S span V?

I know that (t^2 + 1) - (t + 1) = t^2 - t

But I just don't see what that tells me....

2. Oct 24, 2004

### shmoe

This tells you that your vectors are linearly dependant. Do you see why?

V has dimension 3. You have 3 vectors. If they spanned V, then they would be linearly independant as well (it's crucial here that you have 3 vectors and dimV=3, you probably have a theorem to this effect). You've shown they aren't linearly independant, so can they span?

3. Oct 24, 2004

### Spectre5

Thanks for the response....so basically since I reduced the system to 2 equations, it cannot span since it needs 3 equation (vectors) to span it?

Another related question...

let V = vector space of all continous functions on the interval of the real numbers

let S = {sin(t), sin(2t), sin(3t)}

is S of V linearly dependent or lineraly independent?

Thanks any help

4. Oct 24, 2004

### Spectre5

Ok, I get the first question now....since they are linearly dependant, they cannot span a 3d vector space....that makes sense now. I still need some help on that second question though please.

5. Oct 24, 2004

### shmoe

nvm, you got it.

Have you tried the usual test for linear independance? Do you know what it is?

Last edited: Oct 24, 2004
6. Oct 24, 2004

### Spectre5

well, I do know what it is...but this is all I get when using it:

(c1)sin(t) + (c2)sin(2t) + (c3)sin(3t) = 0

I dont know where to go from there

7. Oct 25, 2004

### Spectre5

A set is linearly independant if no vector in the set can be written as a linear combination of of the other vectors

8. Oct 25, 2004

### shmoe

Good. That has to be true for all values of t, so try substituting some in and see what you can say about the coefficients.

Ideally you want to put in a value of t that will make 2 of the terms vanish so you can elimante one coefficient.

E.g. to shaow t and cos(t) are independant, look at

(c1)t+(c2)cos(t)=0 (*)

Put in t=0 and get:

(c1)0+(c2)1=0

so c2=0 and equation (*) becomes

(c1)t=0

Put in t=1, and we see c1=0, therefore t and cos(t) are independant.

9. Oct 25, 2004

### Spectre5

well, I cannot think of any values that would cancel out just two of the constants...which leads me to believe it is linearly independent....but how do I know that there are no values....how do I know I am not missing some value?? Is there a better way of doing it?

10. Oct 25, 2004

### shmoe

It's not necessary to have 2 of them cancel, that would make life easier, but it's not necessary and not always possible.

Putting in any value of t (where they don't all vanish), will allow you to simplify your equation by getting a relation among c1, c2, and c3. Can you find any values of t to make one of them vanish and not the other 2?

11. Oct 25, 2004

### Spectre5

then we have (c1) - (c3) = 0
so (c1) = (c3)

12. Oct 25, 2004

### shmoe

Good. You've simplified down to 2 variables now. Try substituting in some more values of t to see what you can get.

Another thing you can do is differentiate your equation. You'll get a linear combination of cos's. Try putting in convenient values for t and see what you can say about c1 and c2. You could differentiate this again if need be.

13. Oct 25, 2004

### Spectre5

differentiate the equation, and you get all of it in terms of cos....then plug in pi/2 again and only the c2 term is left.

so (c2)*2*cost(2(pi/2)) = 0

So (c2) = 0

And then back to the original eq..

(c1)[sin(t) + sint(3t)] = 0

WE can ignore the (c2) term since it is always 0, but now if t = pi/2, then the equation is true for any value of (c1) and thus any value of (c3)

Last edited: Oct 25, 2004
14. Oct 25, 2004

### shmoe

Ok, c2=0 like you showed, that's good, and you had c1=c3 from earlier post, so you're down to (c1)[sin(t) + sint(3t)] = 0, which must be true for all t.

What value of c1 makes (c1)[sin(t) + sint(3t)] = 0 for all values of t? What if t=pi/6, what can you say about c1?

15. Oct 25, 2004

### Spectre5

then c1 is 0 and c3 must also be 0....ok

It doesn't matter that c1 could be anything when t is pi/2 becuase the equation must hold for all values of t, and that is only possible when c1 is zero.

Therefore, the set is linearly independent.

Thanks for all your help!