Question

Show that the following set of vectors is a basis for P_2 (i.e. the set of all
polynomials of degree 2 or less):

Answer 1

What following set of vectors?

Answer 2

Probably something like {1, x, x^2}

Answer 3

S = {x^2 - x +1, 2*x^2 + x + 1, x^2 + x - 4}

sorry that took a while haha

Answer 4

Oh, not so easy as that.

Answer 5

^ HA HA

Answer 6

Haha nope :P Any taker...? Would be much obliged :D

Answer 7

Ok so lets see. First we have to show that they're linearly independent.

Answer 8

Blargh.. I think that way is harder than needed.

Answer 9

Doing that first is probably what I'm supposed to do, though - part (a) of the same question is on LI, as well... :/

Answer 10

use the fact that there exists an isomorphism between P^2 and R^3.

its easier to prove things in R^3 than P^2.

Answer 11

\[c_1(x^2 - x + 1) + c_2(2x^2 + x + 1) + c_3(x^2 +x - 4) = 0\]\[\implies (c_1 + 2c_2 + c_3)x^2 + (-c_1 + c_2 + c_3)x + (c_1 + c_2 -4c_3) = 0\]

This?

Answer 12

Yeah thats what I had... and I got stuck from there :/

Answer 13

Well from there you have that either all of those coefficients add up to 0

Answer 14

err no either

Answer 15

\[1-x+x^2 \rightarrow (1,-1,1)\]\[1+x+2x^2 \rightarrow (1, 1, 2)\]\[-4+x+x^2 \rightarrow (-4, 1, 1)\]

Proving those three vectors are linearly independent is the same thing as proving those 3 polynomials are linearly independent.

Answer 16

Gives you a new system:
\[c_1 + 2c_2 + c_3 = 0\]\[-c_1 + c_2 + c_3 = 0\]\[c_1 + c_2 -4c_3 = 0\]

Answer 17

Which is what joe is saying but he uses fancy words like isomorphism

Answer 18

sry =/ i didnt know how else to say it >.< there really isnt a "down to earth" term for isomorphism >.<

Answer 19

True, but you can explain it. ;p

Answer 20

Hahaha OK, with you so far :)

Answer 21

Which is what I did above.

Answer 22

Actually I explained all that without knowing that it was called an isomorphism

Answer 23

and only realized after the fact when your vectors came up as my coefficients.

Answer 24

sry sry =/ in my defense, im just realizing right now that the reason you can do that is because an isomorphism exists between P2 and R3.
When i was taught how to do these problems, we werent told that. We were just told, "do it this way"

Answer 25

so my head is trying to catch up to what is really going on. my bad.

Answer 26

lol. No worries. I'm sure I'll learn all about isomorphisms next term

Answer 27

So feldy90, did you solve that new system?

Answer 28

Just about to :)

Answer 29

Erm.... I'm only getting the solution that all coefficients are equal to zero...

Answer 30

thats exactly what you want :)

Answer 31

:)

Answer 32

That's good. It means the only solution is the trivial one. Therefore the vectors are linearly independent.

Answer 33

So that proves they are LI?

Answer 34

Woo! (I haven't gotten something right in quite a while... this is making me happy hahaha)

Answer 35

But how, now, do I prove that they are a basis for P_2??

Answer 36

Well that's actually pretty easy

Answer 37

Or does proving that they are linearly independent also prove they are a basis?

Answer 38

It proves they are a basis. But it doesn't prove that the vector space that their basis spans is all of P_2

Answer 39

For it to span all of P_2 we have to show that any polynomial of the form:
\[\forall (a,b,c) \in R^3, ax^2 + bx + c \] can be formed by a linear combination of those 3 vectors.

Answer 40

So the above means "for all values of a, b, and c in space R^3, ax....etc"? How can you show that? Or do you think the question as written would even require it?

Answer 41

I'm trying to think of the easiest way to show that. Ideally I'd like to find linear combinations of the vectors that result in only one term (for each of the muliplicities of x)

Answer 42

It may not be doable.

Answer 43

We could work in R^3 again. get the inverse of the matrix associated with the basis vectors.

Answer 44

and multiply that with the vector (a,b,c)

Answer 45

i'll do it on paper and see if anything turns up. i dont really know if that will do the job.

Answer 46

Thank you both so much for all of your help :)

Answer 47

this is what i have so far.

Answer 48

Here is an example worked out, and also the general formula:

Answer 49

to be honest, this is too much work imo >.< i dont know if its acceptable, but i would have said something along the lines of, "We have a 3-dimensional space (P2), and 3 linearly independent "vectors" (the polynomials), therefore the polynomials must span all of P2."

Answer 50

Yeah, I wanted to say that. Probably a lot easier. We can show that P_2 is 3 dimensional. We did show that our basis was Linearly independent. Since the dimension of the basis is equal to the dimension of P_2, and since all the basis vectors lie in P_2, The basis must span all of P_2. Must nicer than all that extra work. Especially at 3:00 am ;p

Answer 51

I wish this site allowed the giving of multiple medals haha