What is the Basis and Dimension of a Polynomial Space with a Specific Condition?

[Inirit]

Got another linear space question. I'm getting closer to understanding what's going on, but I'm not there yet.

Homework Statement

Find a basis for the space and determine its dimension.

The space of all polynomials f(t) in P2 such that f(1) = 0.

Homework Equations

The Attempt at a Solution

The dimension is trivial, it's just the number of elements in the basis. It's finding the basis that I am lost with. So far I understand that the general form of P2 is a+bx+cx^2, and that if f(1) = 0, then a+b+c = 0. From here I am not sure how to derive a basis. I expressed each coefficient as a combination of the other two, but I don't know what to do with that or if it even helps. I know that the basis of the space P2 is {1,t,t^2}, but I don't know what to do with that either.

 

[fzero]

The basis elements must also satisfy f(1)=0. Find linearly independent combinations of {1,t,t²} that have that property.

 

[Inirit]

Really? So the independent combinations would be 1-t and 1-t^2, right? Therefore, the basis would be {1-t,1-t^2}, which would have a dimension of 2.

 

[fzero]

Really? So the independent combinations would be 1-t and 1-t^2, right? Therefore, the basis would be {1-t,1-t^2}, which would have a dimension of 2.

Are you convinced that the subspace has dimension 2? Could t² - t be in the basis?

 

[Inirit]

t^2 - t is a linear combination of 1 - t and 1 - t^2 because 1(1 - t) - 1(1 - t^2) = t^2 - t, therefore it's merely within the space and doesn't form part of the basis.

...Right?

 

[Outlined]

hint: by unique factorization of your polynomial for its given zero points your space looks as follows:

\{a(t - 1)(t - b) : a, b \in \textbf{R} \mbox{ where } a \ne 0\} {\cup}<br /> \{a(t - 1): a \in \textbf{R} \mbox{ where } a \ne 0\} {\cup}<br /> \{1\}

[I assume your polynomials are real valued]

 

[fzero]

t^2 - t is a linear combination of 1 - t and 1 - t^2 because 1(1 - t) - 1(1 - t^2) = t^2 - t, therefore it's merely within the space and doesn't form part of the basis.

...Right?

That's right. You can also use the condition a+b+c = 0 that you found before to compute the dimension of the subspace.