Question

Problem 16. Let V be a subspace of R", and suppose that (71, ... , Um) is a basis of V. 1. Show that (v1 – 02, v2 – 03, ..., Um-1 ample. Um, Um – 01) is also a basis of V, or find a counterex- 2. Show that (ủ1 – 02, 02 – 03, ..., Um-1 – Um, Um) is also a basis of V, or find a counterexample.

Answer 1

* solution * 1. let v be a subspace of IRN & suppose that { v. 12. -- Vm } is a basis of v. Then I 33 ...-, Vm-1-Vm Im-V ? may not be basis for v.. * counter Example to Let v= 2(3, 4, 2,O) / 2, 4, ZER} - is a subspace of R. Car let {(1,0,0,0), 0.1, 0, 0), (0040)} is a basis for vi. But {(1,0,0,0)-(0,1,0,0,0,1,0,0)-0,0,1,0), 10040) - (1000) - {(1,1,0,0), (0, 1,-1,0), (-1,0 It is not basis for v.. Because it is linearly dependent. for consider AC1,-1, 0, 0)+bcool,-1,0)+((-1,0,1,0) =(0,0,0,0) (a,-a,0,0) +(0, b, b, o)+(-C, o, c, o) = (0,0,0,0) (a-c, -atb -btc) = (0,0,0). Equating we get a-c=0, -a+6=0,-b+c=0

DATE / /20 = aac, a=b, 6=c.. रंर टव, प, टटय. for a=1 a=6= c = 1. (non zero ) hence {61,-1,0,0), (0, 1,-1,0), (-1,0,1,0)} is linearly dependent let let basis 3 Now v be a swospace of IR". B= {2, D2, D3, . - - , Im} be a for v. dim Vem let B = {1, 12, 12-V3, Van, To show that B is linearly independent set. Consider, X CV, 32 +4 (727) + 2 (3-1) + - - - - + Lm- (Im-Pm) + Lm um = 0 ... Vid V2 + 42 42 - 22 T3 + 2 h + --- -- +Lm-1 Vm-1 - Km 1 Im tam im = 0 KN +(62-21) V2 + (dz-22) Vit. . . . .+(&m-km-1m = 0 since B is Linearly independent Save The Girl Chile

4,20 42-4,50 L3-L220 L=0 93=0 Km-1-Xm-2=0 Lm - Lm 20 km-1=0. km=o. te 4 = x = x3 = - - - =am=o... Thus B' is linearly independent ALSO B contains m no of rectors .. & dim v=m. we have, AA linearly independent set containing in vectors Cdim v) is a basis for v. Hence B is basis for v.