Subspaces of polynomials with degree <= 2

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The subset {p(t): p(0) = 2} is not a subspace of P2 because it does not include the zero polynomial, which is required for any subspace. The zero polynomial, p(t) = 0, must satisfy p(0) = 0, but in this case, all polynomials in the subset evaluate to 2 at t=0. A polynomial in this subset can be expressed as 2 + bt + ct^2, but cannot equal zero for all t. Therefore, since it lacks the neutral element of P2, it cannot be considered a subspace. This understanding clarifies why the subset fails to meet the criteria for a subspace.
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Homework Statement


Which of these subsets of P2 are subspaces of P2? Find a basis for those that are subspaces.

(Only one part)
{p(t): p(0) = 2}

Homework Equations



The Attempt at a Solution


So, I know the answer is that it's not a subspace via back of the book, but I don't understand why. Supposedly it's because it doesn't contain p(t) = 0 for all t, but I don't see why that's true. If p(0) = 2, then the equation has to look like 2 + bt + ct^2. If b,c = -1, then p(1) = 0, therefore there is a way for p(t) = 0.

Maybe I am just not understanding something. Admittedly, I feel really lost with all this.
 
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Yes, there might be a t such that p(t)=0. But we need p(t)=0 for ALL POSSIBLE t.

In your example, we have the poly 2-t-t^2. Then p(1)=0, but that's not good enough. We want p(t)=0 for all t, not just 1.
 
So... the neutral element of P2 has to equal 0 for all t? Therefore, a,b,c would have to all equal 0 to make that true, but that can't be the case because a has to equal 2, meaning it doesn't contain the neutral element and isn't a subspace of P2... right?

If that's the case, then it makes a bit more sense to me now.
 
Yes, this is correct!
 
Awesome, thanks a lot for the help.
 
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