# Basis and co-ordinates with respect to a basis.

• Dec 5th 2010, 06:13 AM
katedew987
Basis and co-ordinates with respect to a basis.
I don't know where to go with this question:

'Let r be a fixed real number. Show that {1, x + r, (x + r)²} is a basis for R[x]2 (that's meant to be a subscript 2) and find the co-ordinates a0 +a1 +a2x² (integers are also meant to be subscript) with respect to this basis.'

• Dec 5th 2010, 07:21 AM
HallsofIvy
Well, what do you know? You are clearly expected here to know what a "basis" is- it must span the space and be independent. Any "vector" in $\displaystyle R[x]_2$, the space of polynomials, with real coefficients, of degree 2 or less, is of the form $\displaystyle a+ bx+ cx^2$. To show that these functions, 1, x+ r, and $\displaystyle (x+ r)^2$ span the space you must show that whatever a, b, and c are, there exist numbers, $\displaystyle \alpha$, $\displaystyle \beta$, and [tex]\gamma[/b] such that $\displaystyle a+ bx+ cx^2= \alpha(1)+ \beta(x+ r)+ \gamma(x+ r)^2$. To show that they are independent, you must show that the only way you can have $\displaystyle \alpha(1)+ \beta(x+ r)+ \gamma(x+ r)^2= 0$, for all x, is to have $\displaystyle \alpha= \beta= \gamma= 0$.
• Dec 5th 2010, 07:26 AM
hatsoff
Well basically you just want to multiply out $\displaystyle (x+r)^2$, and use the result to find $\displaystyle a_0,a_1,a_2$.

We get $\displaystyle (x+r)^2=x^2+2rx+r^2$, so $\displaystyle x^2=(x+r)^2-2r(x+r)+r^2$ and $\displaystyle x=(x+r)-r$.

Therefore $\displaystyle a_0+a_1x+a_2x^2=a_0+a_1(x-r)-a_1r+a_2(x+r)^2-2a_2r(x+r)+a_2r^2$

$\displaystyle =(a_0+a_2r^2)+(a_1-2a_2r)(x-r)+a_2(x+r)^2$

EDIT: Ah, looks like Hallsofivy beat me to it.