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

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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 $p_1,p_3,p_4$

Does the set p_1=1,p_2=1−3x,p_3=x,p_4=x^2p_1=1,p_2=1−3x,p_3=x,p_4=x^2 form a basis for P2P2?