subspaces & dimension

[bcjochim07]

1. The problem statement, all variables and given/known data
Is the collection a subspace of the given vector space? If so what is the dimension?

V={ax^2+bx+c: a=b+c} in P2


2. Relevant equations



3. The attempt at a solution
The first part of the question is pretty straightforward. I just verified closure under addition and scalar multiplication to show that V is indeed a subspace of P2. But I am confused about the second part. I know that the dimension of P2 is 3, so dimV must be < or = 3. I want to say three, but that would mean that V=P2, and I'm not sure that that's true. Any ideas?

[slider142]

You have the condition that a = b + c, so a is constrained. b and c completely specify a. Can you show the dimension explicitly from this?

[bcjochim07]

Hmmm...ok. I think that a basis for V would be {x^2+x, x^2+1}, and therefore the dimension is 2.

[HallsofIvy]

Okay, why do you think that is a basis? Are x^2+ x and x^2+ 1 independent? Do they span V?

[bcjochim07]

Well, {x^2+x, x^2+1} is linearly independent, and I think they span V since it contains all of the different polynomial terms as well as satisfying the restrictions placed on the coefficients. Is that not right?

[matt grime]

Thinking it is not the same as having a proof for it. Forget polynomials - it's just 3-tuples

(a,b,c)

such that a=b+c. Is (1,1,0) and (0,1,1) a basis? I.e. are the linearly independent, and do they span? Even more explicitly, if x(1,1,0)+y(0,1,1)=(0,0,0) what does that say about x and y? And given (a,b,c) in the subspace, can you write it as r(1,1,0)+s(0,1,1) for a suitable r and s?

[HallsofIvy]

Of course, you can use x2, etc. if you want to!:tongue:

To show they are independent, look at the equation \alpha(x^2+ x)+ \beta(x^2+ 1)= 0 (= 0 for all x). What must \alpha and b equal?

Since V consists of all polynomials of the form ax^2+ bx+ c with a= b+c or (b+c)x^2+ bx+ c, to show that these span V, look at the equation \alpha(x^2+ x)+ \beta(x^2+ 1)= (b+c)x^2+ bc+ d. Can that be solved for \alpha and \beta for all b and c?

[bcjochim07]

Yes, I've got the linear independence part, alpha and beta equal 0. So point #1 for a basis is satisfied by B= {x^2+x, x^2+1}

Ok, so for any arbitrary polynomial in V, (b+c)x^2+bx+c, exists in the span of the basis vectors.

so c1(x^2+x) + c2(x^2+1) = (b+c)x^2+bx+c
=c1x^2 +c1x + c2x^2 + c2 = (b+c)x^2 + bx + c

so the resulting system of equations is:
c1+c2 = b + c
c1=b
c2=c and this system is consistent.

So I have shown that my proposed B is indeed a basis for V and that V is a 2 dimensional subspace of P2.

[matt grime]

You've just assumed the answer - you can't write (b+c)x^2+bx+c as an arbitrary vector in V when you're trying to show that such a decomposition exists.

[HallsofIvy]

Yes, I've got the linear independence part, alpha and beta equal 0. So point #1 for a basis is satisfied by B= {x^2+x, x^2+1}

Ok, so for any arbitrary polynomial in V, (b+c)x^2+bx+c, exists in the span of the basis vectors.

so c1(x^2+x) + c2(x^2+1) = (b+c)x^2+bx+c
=c1x^2 +c1x + c2x^2 + c2 = (b+c)x^2 + bx + c

so the resulting system of equations is:
c1+c2 = b + c
c1=b
c2=c and this system is consistent.
That's the wrong way- you have shown that (b+c)x^2+ bx+ c in of the form ax^2+ bx+c with a= b+ c which was obvious from the definition of the subset.

So I have shown that my proposed B is indeed a basis for V and that V is a 2 dimensional subspace of P2.

I said before
Look at the equation \alpha(x^2+ x)+ \beta(x^2+ 1)= (b+c)x^2+ bc+ d. Can that be solved for \alpha and \beta for all b and c?

Do that!