Subspaces & dimension

• bcjochim07
In summary, the homework statement is that V is a subspace of the given vector space. The dimension of P2 is 3, so dimV must be < or = 3. V=P2 would mean that V=P2, and I'm not sure that that's true. The first part of the question is straightforward, verifying closure under addition and scalar multiplication. The second part is more complicated, I'm not sure what the dimension of P2 is. However, I've shown that a basis for V exists which satisfies the condition that a=b+c. This basis is {x^2+x, x^2+1}. The dimension of V isf

Homework Statement

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

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?

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?

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

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

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?

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?

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?

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.

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.

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!