Question

4. Let B = {x6 + 3, x5 + x3 + 1, x4 + x3, x3 + x2} C Pg, where Pg is the polynomials of degree < 8. (a) (2 marks) Explain why B is a linearly independent subset of Pg. (b) (2 marks) Extend B to a basis of Pg by adding only polynomials from the standard basis of Pg.

Answer 1

9 B={ x 6+3,25 + x² + 1, x4x² 3+x²3. consider a lincar combination of veeters in B. For scalars, a, b, c, d E IR, a(x8+ 3) + b(25 + x² + 1) +6C 24+ x) = 0x8toxtt...fo + dC 3+x2) » a (x6) + b 5) + ((24) = 0x8t ...to. + (btc) x + d( x ²) + (3a +b) compaining coefficient of powers of a » azo, bao cao dao. As all scalans are zero, B linearly independent. b) Now we have to extend B to form basis of P8. standard basis of Pq is { 1 X, 82, x3, ..., x 7 x 8 B hay already 4 rectors. Now add rectors { 1, 2, 27, 287 to B. let B = { x +3,25 +3+1, 84+43 x3 + x², 1, 2, 87, x8] Now to rent that B is a basis. consider scalan a 92 r..98 EIR, such that, la (a +3 +9, (2 5 + x² +1) +9 (244²) on 8 tonttranto. tau ² + y² + ar (1) + 96 (2) of 97 (27) + 98 (28) * qq (28) + Q7 (27) + a, (46) + 92 (25) + a, (24) + (azt az) x² + 94 (x ² = 0x toxtt...d oxto. ta6(a) + (30 +92 + as) Compaining coefficient, we get q = Q = ... = 98 = 0. * As all scalars ax zero, B i linearly independent » B formy a basis for Pq.