Can Spanning Sets Help Me Understand Vector Spaces Better?

[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...

 

[shmoe]

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

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

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

V has dimension 3. You have 3 vectors. If they spanned V, then they would be linearly dependent 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 dependent, so can they span?

 

[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 continuous 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

 

[Spectre5]

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

 

[shmoe]

nvm, you got it.

Another related question...

let V = vector space of all continuous 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

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

 

[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 don't know where to go from there

 

[Spectre5]

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

 

[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 independent, 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 independent.

 

[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?

 

[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?

 

[Spectre5]

how about pi/2

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

 

[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.

 

[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)

 

[shmoe]

(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)

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?

 

[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 because 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!