# Finding a basis

• Mar 13th 2011, 03:33 PM
CaughtARattleSnake
Finding a basis
If possible, find a basis n = {n1, n2, n3} of P2(R) such that

[2 + 5x + 4x^2]n = [1,2,3] , [1 + x + x^2]n = [4,1,2] and

[x + x^2]n = [3, -5, 1]

Basically, from what I understand, we have Ax = b.

I can find a basis, for one of these, but not all three.

like, [a,b,c] * [2,5,4]^t = [1,2,3]^t
[d,e,f]
[g,h,i]

0, 1, -1
1, 0, 0
-1, 1, 0 so, [0,1,-1], [1,0,1] [-1,0,0] would be a basis for the first one, but the others... I'm not sure...

Thank you
• Mar 14th 2011, 09:06 AM
FernandoRevilla
Quote:

Originally Posted by CaughtARattleSnake
[2 + 5x + 4x^2]n = [1,2,3]

What does the above equality mean?. Coordinates of $2 + 5x + 4x^2$ on $n$ ?
• Mar 14th 2011, 09:38 AM
FernandoRevilla
If the answer to my question is yes then, denoting:

$B=\{1,x,x^2\},\;B'=\{2 + 5x + 4x^2,1 + x + x^2,x + x^2\}$

you have enough information to find $P,Q\in\mathbb{R}^{3\times 3}$ such that for every $f(x)\in P_2(\mathbb{R})$ :

$[f(x)]_B=P[f(x)]_{B'},\quad[f(x)]_n=Q[f(x)]_{B'}$

This implies $[f(x)]_B=PQ^{-1}[f(x)]_n$ and finding $PQ^{-1}$ , you determine inmediately $n$ .