Does the set #p_1=1,p_2=1−3x,p_3=x,p_4=x^2# form a basis for #P2#?

Question

1684 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-04T00:15:23

See below.

Any #P_2(x)=a x^2+bx+c# can be represented as

#alpha p_1+beta p_2+gammap_3+delta p_4#

or

#a x^2+bx+c=alpha+beta -(3beta-gamma)x+delta x^2#

with

#{(alpha+beta=a),(-3beta+gamma=b),(delta=c):}#

or

#((0,0,0,1),(0,-3,1,0),(1,1,0,0))((alpha),(beta),(gamma),(delta))=((a),(b),(c))#

or

#((0,0,1),(0,-3,0),(1,1,0))((alpha),(beta),(delta))=((a),(b),(c))-gamma((0),(1),(0))#

Here #gamma# is a free parameter and can be set to #0#. Also

#((0,0,1),(0,-3,0),(1,1,0))# is invertible so a minimal basis is formed by

#p_1,p_2,p_4# or #p_1,p_3,p_4#