Show that the set is a basis for P2R Show that 1 x2 x2 x

Show that the set is a basis for P2(R)

Show that {1 - x^2; x^2 + x + 2; 2x - 3} is a basis for P_2(R).

Solution

It is apparent that 1 –x² , x² + x + 2 and 2x -3 are linearly independent, as none of the three terms can be expressed as a linear combination of the other two terms.An arbitrary elemet of P₂(R) is ax² + bx +c where a.b, c are arbitrary real numbers. Let ax² + bx +c = p(1 –x²) + q(x² + x + 2) + r(2x -3 ) = (q-p)x² + (q+2r)x + (p +2q – 3r). Then a = q - p, b =q +2r nd c = p +2q -3r. On solving these equations, we get p =(-7a+ 3b + 2c)/9, q = (2a+3b +2c)/9 and r = (-a +3b-c)/9. Thus, ax² + bx +c = [(-7a+ 3b + 2c)/9](1 –x²) + [(2a+3b +2c)/9 ]( x² + x + 2) + [(-a +3b-c)/9]( 2x -3). Thus every element of P₂(R) can be expressed as a linear combination of 1 –x² , x² + x + 2 and 2x -3. Therefore, 1 –x² , x² + x + 2 and 2x -3 form a basis for P₂(R)