Basis and Dimension of P2

  • Thread starter Hockeystar
  • Start date
1. The problem statement, all variables and given/known data
Find the basis and dimension of the following subspace U of P2

p(x) [tex]\ni[/tex] P2 such that p(1) = p(2)


2. Relevant equations



3. The attempt at a solution

I know all quadratics are in the form ax2 + bx + c

set p(2) = p(1)

4a + 2b + c = a + b + c
b = -3a

Therefore ax2 -3a + c

Basis(U) = a(x2-3x) + c

Therefore dim(U) = 2

I'm just wondering if I have the correct answer or not. Going into linear algebra midterm tomorrow and prof never really went over polynomials but it's on the test.
 
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I assume P2 is the space of all second degree polynomials (the notation P2 is not standard, so includi it next time :smile: ). Then your proof is correcT.

The basis for U is by the way [tex] \{x^2-3x,1\} [/tex]. What you wrote down doesn't make much sense to me...
 
Thanks for the reply. I just wanted clarification that I'm properly solving the problem. My format was off because I used subscript instead of superscript.
 

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