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anonymous
 5 years ago
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(2x) }. Find a basis for P2(R) that contains a basis for W.
anonymous
 5 years ago
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(2x) }. Find a basis for P2(R) that contains a basis for W.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0p(x)=p(2x) 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 \[ax^2+bx+c=p(x)=p(2x)=a(2x)^2+b(2x)+c.\] Expand both sides and you'll find that \[\begin{array}{rcl} b&=&4ab\\ c&=&4a+2b+c\\ \end{array}\] The solution: \[a\in\mathbb{R},\quad b=2a,\quad c\in\mathbb{R}.\] So dim(W)=2. A basis for W is: \[e_1(x)=1,\quad e_2(x)=x^22x.\] To extend these two vectors to a basis of P2(R) you'd just need to use the vector \[e_3(x)=x.\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great 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??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It' 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 \[e_1=1,\quad e_2=x+2,\quad e_3=x^22x+4\]
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