Question

15 points) Consider the following vectors in R3 0 0 2 V1 = 1 ; V2 = 3 ; V3 = 1] ; V4 = -1;V5 = 4 1 2 3 = a) Are V1, V2, V3, V4, V5 linearly independent? Explain. b) Let H (V1, V2, V3, V4, V5) be a 3 x 5 matrix, find (i) a basis of N(H) (ii) a basis of R(H) (iii) a basis of C(H) (iv) the rank of H (v) the nullity of HT

Answer 1

Given: The following vectors in R² ū VE Ma- (a) Dimension of R²-3 number of vectors = 5 We know that a set containing more than n vectors in an more than n vectors in an n-dimensional vector space cannot be linearly independent. V. { }, U is are not linearly independent. - -2 13 - 2 -1 - 3 2 4 3 2 - 2 2 -1 R & Rz 1 3 3 4 -2 - - 1 L-2 O 2 1 2 2 o 1 - R₂ R₂R, R (+2R, - 4 4 0 4 0 4 -1 o -- -- O R,R, -2R2 Ry R-4R2 o 8 D 1 o 4 -- 1 Rzut R3 1 -1 0 - o 1 0 - 1 -- R,R, 4R3 R₂ R₂ + Rz o 0 Reduced - 0 0 you echelon form, • Let V = m ENCH) us Then, Hu= The augmented matrix is [H10] -2 0 -2 3 4 13 2 2 Nī 7 Too 100 1 0 1 0 - 0 Reduced row echelon form O

- , -y +45 og X₂+ "s=0 =o. n = 44 45 "₂ = -ts "=0 7 - nens ang ny us 0 N {930 A basis for NCH) = ii) All the three rows in the reduced ww. echelon form, so they form a basis for R(H) :. A basis for R(H) = {(2,0,0,2,-2), (1,3 1,-1,4),(1,2,3,43)} till the pivot elements occur in the first second, third columns in the reduced row echelon form. the first second third column rectors are linearly independent and form a basis for CCH) . A basis for CCH) = {D (120

(iv) Rank of H= number of non-zero rows in sref of H -3 (V) Rank of HT= Rank of H = 3 ..By the rank-nullity theorem rank of His nullity of HT = number of columns in Ho 3+ nullity of HT=3 hullity of HT - 3-3=0