Question

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.

Answer 1

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
\[e_3(x)=x.\]

Answer 2

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??

Answer 3

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
\[e_1=1,\quad e_2=-x+2,\quad e_3=x^2-2x+4\]