Polynomial basis proof

[boneill3]

Hi guy's I was wondering if someone could give me some tips in proving that:

{1,t,t^2} is a basis for real polynomials P^2

I know I need to show 2 things,

1 That the vectors are linearly indepenedent and
2 They generate all vectors in P^2

Suppose we have real numbers a b c

I need to show that


a(1) + b(t) + c(t^2) = 0


but not sure how to prove this



To show that it generates polynomials in P^2 can I just choose a polynomial of P^2 and show the linear combination?

eg

To generate 5+2t +6t^2

we let e_1 = 1 e_2 = t and e_3 = t^2

5+2t +6t^2 = 5*(e_1)+ 2*(e_2) + 6*(e_3)

regards
Brendan

[boneill3]

I put the set {1,t,t^2} into matrix form we produced the natural basis for dim 3

1 0 0
0 1 0
0 0 1


As P^2 has dimention n+1 = 3

The set is dim 3

therefore the set {1,t,t^2} is a basis

As the basis is the standard basis do I have to still prove:

2) They generate all vectors in P^2
regards
Brendan

[jemidiah]

Any basis of a vector space spans that vector space, meaning that linear combinations of any basis yield all vectors of that vector space.

I'm not convinced you've shown {1, t, t^2} is a basis. I could try the "basis" {1+t, t^2, 1+t+t^2} and put the vectors 1+t, t^2, and 1+t+t^2 into an identity matrix too. It doesn't mean those vectors span the set. (They don't.)


What you need to show is that linear combinations of {1, t, t^2} generate all elements of P^2. What are the elements of P^2? Polynomials of the form a+bt+ct^2, by definition. Does your set span these elements? If it does, then if I gave you a specific polynomial in the set by specifying a, b, and c, then you would be able to give me a linear combination of your basis that equals that polynomial. You did this in a specific case in your first post. If you can do it in general (it's quite easy here), then you've proven that your set spans your vector space.

However, the above doesn't answer linear independence. For instance, {1, 2, t, t^2} spans P^2. To show linear independence, you need to do one of two things: (1) if you know the dimension of the vector space you're working in, show that your set has size equal to that dimension and spans the space; (2) if you don't know the dimension of your vector space, show that linear combinations of your basis elements are zero only when the coefficients are *all* zero.

That is, take a1*1 + a2*t + a3*t^2 and show a1*1 + a2*t + a3*t^2 = 0 only when a1=a2=a3=0. There's a somewhat subtle point that might creep up here--the "0" on the right hand side is actually a polynomial in the set P^2. That is, the "0" is a formal symbol standing in for the zero function, that is, the map that takes in one argument (t) and returns 0 all the time. To show this, think about what happens when one of the ai's are non-zero--is the resulting sum always zero then? Is it even constant? How often is it zero? You should be able to say pretty quickly that (with one of the ai's non-zero) the expression on the left cannot be the zero function.

[boneill3]

To prove that the basis span the set

We take a,b,c all elements of R

than for any polynomial in P^2 in the form

a*1 + b*t + c*t^2 = a*(e_1)+ b*(e_2) + c*(e_3)

As for the linearly indepenedence as the dimension of P_n = P_n+1 = 2 + 1 = 3 and the set {1,t,t^2} is dimension 3

As for the linearly independence
could we have a prrof by contradiction?

say we asume there is no combination of vectors that can produce the zero vector.

and then show thers is a least one.

ie

using a,b,c all elements of R.

a = 0 and
b=c not equal to a

a*1 + b*t + c*t^2 = bt+ct^2 not= 0


b=0=c
a not = b or c

a*1 + b*t + c*t^2 = a not= 0



c=0 not equal to a = b


a*1 + b*t + c*t^2 = a +bt not= 0


c=0 = a = b

a*1 + b*t + c*t^2 = 0 + 0 + 0 = 0

Proving there is a least one proving linear indepenedence bu contradiction

[jemidiah]

To prove that the basis span the set

We take a,b,c all elements of R

than for any polynomial in P^2 in the form

a*1 + b*t + c*t^2 = a*(e_1)+ b*(e_2) + c*(e_3)

As for the linearly indepenedence as the dimension of P_n = P_n+1 = 2 + 1 = 3 and the set {1,t,t^2} is dimension 3

You're good through here.

As for the linearly independence
could we have a prrof by contradiction?

say we asume there is no combination of vectors that can produce the zero vector.

and then show thers is a least one.

ie

using a,b,c all elements of R.

a = 0 and
b=c not equal to a

a*1 + b*t + c*t^2 = bt+ct^2 not= 0


b=0=c
a not = b or c

a*1 + b*t + c*t^2 = a not= 0



c=0 not equal to a = b


a*1 + b*t + c*t^2 = a +bt not= 0


c=0 = a = b

a*1 + b*t + c*t^2 = 0 + 0 + 0 = 0

Proving there is a least one proving linear indepenedence bu contradiction

This is a bit of a tangled mess. You're assuming, to the contrary, that for some a, b, and c (not all zero), that the linear combination a*1+b*t+c*t^2 = 0--so far so good. You then cycle through many of the possibilities--a=0, b and c non-zero; b and c zero, a non-zero; a, b, and c all zero (I don't know why this case is in here, actually...). Regardless, your list of possibilities isn't exhaustive.

What you would want to do (pursuing your line of reasoning; of course more elegant proofs exist) is an exhaustive case-by-case analysis, like the one I'll start below:

Assume, to the contrary, that some a, b, and c (not all zero) exist, such that a+bt+ct^2=0. Then...

If a is non-zero, then for t=0, a+bt+ct^2 = a <> 0, a contradiction. Thus a is zero. Since a is zero, either b or c is non-zero. If c is zero, then b must be non-zero--but then at t=1 we again have a+bt+ct^2 = b <> 0; thus c is non-zero. If b is zero, the same type of contradiction occurs, so that b is non-zero. Thus, both b and c is non-zero. However, at t = <something>, we see that the polynomial is non-zero <insert specifics here>, a contradiction. Thus, no such a, b, and c, exist, forcing {1, t, t^2} to be linearly independent.


I've of course left the critical step for you to do.

[boneill3]

Are you talking about the negative reals?

say
However, at t = -1, with a non zero we see that the polynomial is non-zero a+b(-1)+c(-1^2) = a <> 0, a contradiction. Thus, no such a, b, and c, exist, forcing {1, t, t^2} to be linearly independent.


regards

[jemidiah]

Are you talking about the negative reals?

say
However, at t = -1, with a non zero we see that the polynomial is non-zero a+b(-1)+c(-1^2) = a <> 0, a contradiction. Thus, no such a, b, and c, exist, forcing {1, t, t^2} to be linearly independent.


regards

Not really... I'm not sure where you got a + b(-1) + c (-1^2) = a from. This expression is just a - b + c--and who knows if that's zero? It depends on which a, b, and c we choose. You can certainly narrow down the possibilities for a, b, and c from this, but not much more.

If you could tell me where you're stuck maybe that'd be more helpful. From your post at least, it looks like you're not being systematic enough--you're going with whatever ideas occur to you without arranging them neatly and rigorously. But, I could be totally off.