Linear Algebra: Let P2(R) denote the vector space of polynomials of degree less than or equal to 2. Let W = { p(x) in P2(R) | p(x) = p(2-x) }. Find a basis for P2(R) that contains a basis for W.

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Linear Algebra: Let P2(R) denote the vector space of polynomials of degree less than or equal to 2. Let W = { p(x) in P2(R) | p(x) = p(2-x) }. Find a basis for P2(R) that contains a basis for W.

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p(x)=p(2-x) We can write an arbitrary element of P2(R) in the form \[p(x)=ax^2+bx+c\textrm{, where }a,b,c\in\mathbb{R}.\] The condition for the element of W is \[ax^2+bx+c=p(x)=p(2-x)=a(2-x)^2+b(2-x)+c.\] Expand both sides and you'll find that \[\begin{array}{rcl} b&=&-4a-b\\ c&=&4a+2b+c\\ \end{array}\] The solution: \[a\in\mathbb{R},\quad b=-2a,\quad c\in\mathbb{R}.\] So dim(W)=2. A basis for W is: \[e_1(x)=1,\quad e_2(x)=x^2-2x.\] To extend these two vectors to a basis of P2(R) you'd just need to use the vector \[e_3(x)=x.\]
great answer thanks! but I got the basis to be : \[e _{1} = 1 , e _{2} = -x + 2 , e _{3} = x^2 - 4x + 4 \] Does that look right??
It' almost OK, there is only one problem. In e_3 the coefficient of x^2 is 1 and because of that the coefficient of x must be -2 NOT -4. So your basis should be \[e_1=1,\quad e_2=-x+2,\quad e_3=x^2-2x+4\]

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